Unix timestamp to datetime string

Given a unix timestamp as an input, give a datetime string, in a format like so: "YYMMDD.HHmm"

Rules

The input is a number (integer) of a millisecond-precise UNIX epoch time (milliseconds since 1970 January 1st 00:00:00.000 UTC).

The values must be padded with zeroes if they are 1 character instead of 2. (e.g.: for "DD", "1" is not acceptable, but "01" is.)

The output must be a single string. No arrays.

Leap second handling doesn't matter.

Shortest wins.

Good luck!

Example

Input: 1547233866744

Output: 190111.1911

edited Jan 12 at 21:57

Community♦

asked Jan 11 at 19:27

skiilaa

1638

2

$begingroup$
Welcome to PPCG! Nice challenge, but there are a few clarifications to be made. To make the challenge self-contained, you should probably explain what the Unix epoch is. Additionally, what should be done with leap seconds? What should be done with the Year 2038 problem?
$endgroup$
– AdmBorkBork
Jan 11 at 19:43

1

$begingroup$
@AdmBorkBork Edited the post to explain what the Unix epoch is. Leap second implementation doesn't matter, since the output string is not second-precise. The Year 2038 problem doesn't currently matter since it could be a limitation of the running device or the chosen programming language.
$endgroup$
– skiilaa
Jan 11 at 19:50

1

$begingroup$
Does the timezone matter?
$endgroup$
– Erik the Outgolfer
Jan 11 at 21:52

$begingroup$
I take it the year is represented by a two digit number?
$endgroup$
– Embodiment of Ignorance
Jan 11 at 22:22

1

$begingroup$
@FabianRöling timezones don't matter.
$endgroup$
– skiilaa
Jan 12 at 18:04

|
show 12 more comments

Given a unix timestamp as an input, give a datetime string, in a format like so: "YYMMDD.HHmm"

Rules

The input is a number (integer) of a millisecond-precise UNIX epoch time (milliseconds since 1970 January 1st 00:00:00.000 UTC).

The values must be padded with zeroes if they are 1 character instead of 2. (e.g.: for "DD", "1" is not acceptable, but "01" is.)

The output must be a single string. No arrays.

Leap second handling doesn't matter.

Shortest wins.

Good luck!

Example

Input: 1547233866744

Output: 190111.1911

edited Jan 12 at 21:57

Community♦

asked Jan 11 at 19:27

skiilaa

1638

2

$begingroup$
Welcome to PPCG! Nice challenge, but there are a few clarifications to be made. To make the challenge self-contained, you should probably explain what the Unix epoch is. Additionally, what should be done with leap seconds? What should be done with the Year 2038 problem?
$endgroup$
– AdmBorkBork
Jan 11 at 19:43

1

$begingroup$
@AdmBorkBork Edited the post to explain what the Unix epoch is. Leap second implementation doesn't matter, since the output string is not second-precise. The Year 2038 problem doesn't currently matter since it could be a limitation of the running device or the chosen programming language.
$endgroup$
– skiilaa
Jan 11 at 19:50

1

$begingroup$
Does the timezone matter?
$endgroup$
– Erik the Outgolfer
Jan 11 at 21:52

$begingroup$
I take it the year is represented by a two digit number?
$endgroup$
– Embodiment of Ignorance
Jan 11 at 22:22

1

$begingroup$
@FabianRöling timezones don't matter.
$endgroup$
– skiilaa
Jan 12 at 18:04

|
show 12 more comments

Given a unix timestamp as an input, give a datetime string, in a format like so: "YYMMDD.HHmm"

Rules

The input is a number (integer) of a millisecond-precise UNIX epoch time (milliseconds since 1970 January 1st 00:00:00.000 UTC).

The values must be padded with zeroes if they are 1 character instead of 2. (e.g.: for "DD", "1" is not acceptable, but "01" is.)

The output must be a single string. No arrays.

Leap second handling doesn't matter.

Shortest wins.

Good luck!

Example

Input: 1547233866744

Output: 190111.1911

edited Jan 12 at 21:57

Community♦

asked Jan 11 at 19:27

skiilaa

1638

Given a unix timestamp as an input, give a datetime string, in a format like so: "YYMMDD.HHmm"

Rules

The input is a number (integer) of a millisecond-precise UNIX epoch time (milliseconds since 1970 January 1st 00:00:00.000 UTC).

The values must be padded with zeroes if they are 1 character instead of 2. (e.g.: for "DD", "1" is not acceptable, but "01" is.)

The output must be a single string. No arrays.

Leap second handling doesn't matter.

Shortest wins.

Good luck!

Example

Input: 1547233866744

Output: 190111.1911

code-golf date

edited Jan 12 at 21:57

Community♦

asked Jan 11 at 19:27

skiilaa

1638

edited Jan 12 at 21:57

Community♦

asked Jan 11 at 19:27

skiilaa

1638

edited Jan 12 at 21:57

Community♦

edited Jan 12 at 21:57

Community♦

edited Jan 12 at 21:57

Community♦

asked Jan 11 at 19:27

skiilaa

1638

asked Jan 11 at 19:27

skiilaa

1638

asked Jan 11 at 19:27

skiilaa

1638

2

$begingroup$
Welcome to PPCG! Nice challenge, but there are a few clarifications to be made. To make the challenge self-contained, you should probably explain what the Unix epoch is. Additionally, what should be done with leap seconds? What should be done with the Year 2038 problem?
$endgroup$
– AdmBorkBork
Jan 11 at 19:43

1

$begingroup$
@AdmBorkBork Edited the post to explain what the Unix epoch is. Leap second implementation doesn't matter, since the output string is not second-precise. The Year 2038 problem doesn't currently matter since it could be a limitation of the running device or the chosen programming language.
$endgroup$
– skiilaa
Jan 11 at 19:50

1

$begingroup$
Does the timezone matter?
$endgroup$
– Erik the Outgolfer
Jan 11 at 21:52

$begingroup$
I take it the year is represented by a two digit number?
$endgroup$
– Embodiment of Ignorance
Jan 11 at 22:22

1

$begingroup$
@FabianRöling timezones don't matter.
$endgroup$
– skiilaa
Jan 12 at 18:04

|
show 12 more comments

2

$begingroup$
Welcome to PPCG! Nice challenge, but there are a few clarifications to be made. To make the challenge self-contained, you should probably explain what the Unix epoch is. Additionally, what should be done with leap seconds? What should be done with the Year 2038 problem?
$endgroup$
– AdmBorkBork
Jan 11 at 19:43

1

$begingroup$
@AdmBorkBork Edited the post to explain what the Unix epoch is. Leap second implementation doesn't matter, since the output string is not second-precise. The Year 2038 problem doesn't currently matter since it could be a limitation of the running device or the chosen programming language.
$endgroup$
– skiilaa
Jan 11 at 19:50

1

$begingroup$
Does the timezone matter?
$endgroup$
– Erik the Outgolfer
Jan 11 at 21:52

$begingroup$
I take it the year is represented by a two digit number?
$endgroup$
– Embodiment of Ignorance
Jan 11 at 22:22

1

$begingroup$
@FabianRöling timezones don't matter.
$endgroup$
– skiilaa
Jan 12 at 18:04

Welcome to PPCG! Nice challenge, but there are a few clarifications to be made. To make the challenge self-contained, you should probably explain what the Unix epoch is. Additionally, what should be done with leap seconds? What should be done with the Year 2038 problem?

– AdmBorkBork
Jan 11 at 19:43

@AdmBorkBork Edited the post to explain what the Unix epoch is. Leap second implementation doesn't matter, since the output string is not second-precise. The Year 2038 problem doesn't currently matter since it could be a limitation of the running device or the chosen programming language.

– skiilaa
Jan 11 at 19:50

Does the timezone matter?

– Erik the Outgolfer
Jan 11 at 21:52

I take it the year is represented by a two digit number?

– Embodiment of Ignorance
Jan 11 at 22:22

@FabianRöling timezones don't matter.

– skiilaa
Jan 12 at 18:04

|
show 12 more comments

21 Answers
21

active

oldest

votes

Japt v1.4.5, 19 16 bytes

GîÐU s3)¤o>J i.G

1 byte saved thanks to Oliver, which led to another 2 bytes saved.

Try it

Explanation

GîÐU s3)¤o>J i.G

                     :Implicit input of integer U

G                    :16

 î                   :Get the first 16 characters of the following string

  ÐU                 :  Convert U to a date object

     s3              :  Convert to ISO string

       )             :End get

        ¤            :Slice off the first 2 characters

         o           :Filter

          >J         :  Greater than -1

             i.G     :Insert "." at 0-based index 16, with wrapping

Notes / Tips

Getting the first 16 characters of the ISO string and then slicing off the first two is a byte shorter than performing a single slice.

G is used to insert the . at the required index because using a literal 6 would cause it to be combined with the . and for that to be interpreted as a decimal that would be inserted at the start of the string. To get around that I'd need to add a ' before the . to force it to be interpreted as a string.

Japt v1.4.5 is used because Japt doesn't have a constant for 17 and from v1.4.6 on trying to insert something at the first index past the end of a string results in it being inserted at the end of the string (A is the Japt constant for 10) whereas prior to v1.4.6 it immediately wraps back to the beginning of the string.

edited Jan 14 at 18:44

answered Jan 11 at 22:34

Shaggy

19.5k21666

add a comment |

JavaScript (ES6), 65 bytes

n=>'101010.1010'.replace(i=/d/g,x=>new Date(n).toJSON()[i=x-~i])

Try it online!

How?

We initialize the pointer $i$ to a non-numeric value (coerced to $0$) and then add alternately $2$ and $1$ to it to pick the relevant characters from the ISO-8601 conversion of the input timestamp.

yyyy-mm-ddThh:mm:ss.sssZ

  ^^ ^^ ^^ ^^ ^^

JavaScript (ES6), 66 bytes

n=>'235689.BCEF'.replace(/w/g,x=>new Date(n).toJSON()[+`0x${x}`])

Try it online!

How?

Once the input timestamp is converted in ISO-8601 format, all required characters can be accessed with a single hexadecimal digit.

yyyy-mm-ddThh:mm:ss.sssZ

  ↓↓ ↓↓ ↓↓ ↓↓ ↓↓

0123456789ABCDEF

edited Jan 12 at 20:49

answered Jan 12 at 10:51

Arnauld

74.1k690310

$begingroup$
Wow. Just, wow.
$endgroup$
– skiilaa
Jan 13 at 9:08

1

$begingroup$
64 bytes: n=>'235689.11121415'.replace(/1?w/g,x=>new Date(n).toJSON()[x])
$endgroup$
– tsh
Jan 14 at 7:02

$begingroup$
@tsh Nice! I was not expecting the decimal format to be shorter. You may want to post this as a separate answer.
$endgroup$
– Arnauld
Jan 14 at 15:45

add a comment |

Bash + coreutils, 29 bytes

date -d@${1::-3} +%y%m%d.%H%M

Try it online!

answered Jan 11 at 19:59

Neil

80k744178

add a comment |

PHP, 40 32 31 bytes

-8 bytes thanks to Luis felipe
-1 byte thanks to Jo King

<?=date('ymd.hi',$argv[1]/1e3);

Try it online!

Simple naive answer. PHP's date function takes a format string and an integer timestamp. Input from cli arguments, which is a string by default, then /1e3 because date expects second-precise timestamps. This also coerces the string to a number.

edited Jan 14 at 14:02

answered Jan 11 at 20:00

Skidsdev

6,4262974

$begingroup$
Nice answer! It's a tad shorter if the language has date formatting, yes :)
$endgroup$
– skiilaa
Jan 11 at 20:03

$begingroup$
32 bytes you dont need to cast to integer since "2"/1 will cast automatically the string
$endgroup$
– Luis felipe De jesus Munoz
Jan 11 at 20:04

$begingroup$
ah of course, I added intval before realising I needed /1000, and didn't think that I mihgt not need it afterwards :P
$endgroup$
– Skidsdev
Jan 11 at 20:08

$begingroup$
@JoKing ah nice one, thanks
$endgroup$
– Skidsdev
Jan 14 at 14:03

add a comment |

MATL, 28 bytes

Thanks to @skiilaa for a correction in the output format.

864e5/719529+'YYmmDD.HHMM'XO

Try it online!

Explanation

MATL, like MATLAB, defines date/time numbers as the (possibly non-integer) number of days since time 00:00 of the reference "date" 0-Jan-0000.

Thus we take the input, divide it by 86400000 (number of milliseconds in one day), add 719529 (number of days from MATL's reference to UNIX reference), and convert to the desired format 'YYmmDD.HHMM'.

edited Jan 13 at 16:06

answered Jan 11 at 22:00

Luis Mendo

74.2k887291

add a comment |

GNU AWK, 34 33 characters

$0=strftime("%y%m%d.%H%M",$0/1e3)

(strftime() is GNU extension, will not run in other AWK implementations.)

Thanks to:

Jo King for suggesting E-notation (-1 character)

Sampler run:

bash-4.4$ awk '$0=strftime("%y%m%d.%H%M",$0/1e3)' <<< 1547233866744

190111.2111

Try it online!

edited Jan 13 at 16:05

answered Jan 11 at 20:28

manatwork

16.3k43572

$begingroup$
Thank you, @JoKing. Unfortunately E-notation is not really my friend so I always forget it.
$endgroup$
– manatwork
Jan 13 at 16:02

add a comment |

PowerShell, 59 58 bytes

"{0:yyMMdd.HHmm}"-f(Date 1/1/1970).AddSeconds("$args"/1e3)

Try it online!

Gets the Date of 1/1/1970 (defaults to 00:00:00am), then Adds the appropriate number of Seconds. Passes that to the -format operator, which correctly formats the datetime.

Probably culture-dependent. This works on TIO, which is en-us.

-1 byte thanks to shaggy.

edited Jan 14 at 13:39

answered Jan 11 at 20:15

AdmBorkBork

26.6k364229

add a comment |

Perl 6, 111 89 87 bytes

{~DateTime.new($_/Ⅿ,:formatter{"{(.year%Ⅽ,.month,.day).fmt('%02d','')}.{(.hour,.minute).fmt('%02d','')}"})}

Try it (111)

{TR/-//}o{S/..//}o{.yyyy-mm-dd~'.'~(.hour,.minute).fmt('%02d','')}o{DateTime.new($_/Ⅿ)}

Try it (89)

{TR/- //}o{S/..//}o{.yyyy-mm-dd~'.'~(.hour,.minute).fmt('%02d')}o{DateTime.new($_/Ⅿ)}

Try it (87)

Explanation:

The o infix operator takes two functions and creates a composite function. The rightmost one gets called first, and the one to the left gets called with its result.

Basically we use 4 block lambdas to generate a single lambda.

Which is not much different to how a WhateverCode lambda like * + * gets created.

Divide by 1000 and use that to create a DateTime object.

{DateTime.new($_/Ⅿ)} # Ⅿ is ROMAN NUMERAL ONE THOUSAND (3 bytes)

The result gets used by:

{

   .yyyy-mm-dd # 2019-01-11



   ~ '.' ~     # str concatenation with '.'



   ( .hour, .minute ).fmt('%02d') # add leading 0s (returns List)

}

That leaves us with a string like 2019-01-11.19 11

We need to remove the first two digits

{S/..//}

We also need to remove - and

{TR/- //}

edited Jan 11 at 21:37

answered Jan 11 at 20:46

Brad Gilbert b2gills

12.2k11232

add a comment |

Python 2, 64 bytes

lambda s:strftime('%y%m%d.%H%M',gmtime(s/1e3))

from time import*

Try it online!

The input is considered to be in UTC.

edited Jan 11 at 22:01

answered Jan 11 at 21:55

Erik the Outgolfer

31.6k429103

1

$begingroup$
Just as something I've been a bit confused on, is the only reason that the lambda comes before the import because it works better for TIO's header?
$endgroup$
– Neil A.
Jan 11 at 22:43

2

$begingroup$
@NeilA. Yes. The import can go either before or after, the contents of the lambda aren't checked for NameErrors before it's called.
$endgroup$
– Erik the Outgolfer
Jan 11 at 22:45

add a comment |

R, 58 56 bytes

format(as.POSIXct(scan()/1e3,,'1970-1-1'),'%y%m%d.%H%M')

Try it online!

edited Jan 12 at 12:03

answered Jan 12 at 11:22

digEmAll

3,029414

add a comment |

C (gcc) (32-bit, little endian), 67 bytes

f(t,s)long long t;{t/=1e3;strftime(s,12,"%y%m%d.%H%M",gmtime(&t));}

Try it online!

On an ILP64 platform, the following 55 byte version should work:

f(t,s){t/=1e3;strftime(s,12,"%y%m%d.%H%M",gmtime(&t));}

edited Jan 12 at 15:08

answered Jan 12 at 14:32

nwellnhof

6,67511126

$begingroup$
What's the extra s argument you're taking for?
$endgroup$
– Shaggy
Jan 12 at 14:47

1

$begingroup$
@Shaggy The s is for the output string.
$endgroup$
– nwellnhof
Jan 12 at 15:06

$begingroup$
It looks like you're initiating it outside the function; do we have a meta consensus that allows C to do that?
$endgroup$
– Shaggy
Jan 12 at 15:31

$begingroup$
@Shaggy I'm not sure what "consensus" means exactly but here's the relevant meta post.
$endgroup$
– nwellnhof
Jan 12 at 15:54

$begingroup$
With "32-bit,", why long long instead of long or int32_t, int64_t, time_t? IAC, all shorter than long long.
$endgroup$
– chux
Jan 13 at 22:23

|
show 2 more comments

Perl 6, 57 50 bytes

{TR:d/T:-/./}o{substr ~DateTime.new($_/1e3): 2,15}

Try it online!

Takes the default stringification of a Datetime, in the format yyyy-mm-ddThh:mm:ssZ and modifies it to fit the output format. Perl 6 is in need of a date formatter method.

Explanation:

                       Datetime.new($_/1e3) # Create a date time

                      ~                 # Stringify it to the format yyyy-mm-ddThh:mm:ssZ

                                        # e.g. 2019-01-11T19:11:06.744000Z

               substr                      : 2,15  # Take the middle 15 characters

 {TR:d/T  /./}o   # Then replace 'T' with '.'

        :-        # Then remove ':' and '-'

edited Jan 11 at 23:38

answered Jan 11 at 22:50

Jo King

21.5k248110

add a comment |

C# (Visual C# Interactive Compiler), 67 61 60 bytes

n=>$"{new DateTime(1970,1,1).AddTicks(n*10000):yyMMdd.HHmm}"

For reasons unknown to me, DateTime.UnixEpoch doesn't work.

Try it online!

edited Jan 12 at 4:11

answered Jan 11 at 22:41

Embodiment of Ignorance

721116

1

$begingroup$
It seems UnixEpoch is only present in.Net Core 2.1+
$endgroup$
– digEmAll
Jan 12 at 12:19

add a comment |

Javascript ES6, 76 66 bytes

x=>new Date(x).toJSON().slice(2,16).replace(/D/g,a=>a>'S'?'.':'')

Try it online

-10 bytes thanks to Shaggy!

x // timestamp

=>

new Date(x) // date object from timestamp

.toJSON() // same as .toISOString()

.slice(2,16) // cut off excess

.replace(/D/g, // match all non-digits

a // a is matched character

=>

a>'S'?'.' // if a is T (bigger than S is shorter) replace it with .

:''       // if it's not T, replace it with nothing

          // this way the dashes get removed and the dot gets put in the right place

) // end of replace

edited Jan 12 at 18:38

answered Jan 11 at 19:27

skiilaa

1638

3

$begingroup$
You may want to wait a day or so before answering your own questions next time.
$endgroup$
– fəˈnɛtɪk
Jan 11 at 20:11

$begingroup$
71 bytes
$endgroup$
– Luis felipe De jesus Munoz
Jan 11 at 20:16

$begingroup$
Alternative 71 bytes
$endgroup$
– Shaggy
Jan 11 at 20:33

1

$begingroup$
@LuisfelipeDejesusMunoz, that's different enough for you to post yourself.
$endgroup$
– Shaggy
Jan 11 at 21:49

$begingroup$
@LuisfelipeDejesusMunoz 66 bytes to tie you with Arnauld.
$endgroup$
– Shaggy
Jan 12 at 13:50

add a comment |

C (clang), 117 111 bytes

Thanks to @chux and @ceilingcat for the suggestions.

#import<time.h>

*l;f(long t){t/=1e3;printf("%02d%02d%02d.%02d%02d",5[l=gmtime(&t)]%100,l[4]+1,l[3],l[2],l[1]);}

Try it online!

edited Jan 14 at 12:15

answered Jan 12 at 3:26

ErikF

1,29917

$begingroup$
gmtime is short than localtime
$endgroup$
– chux
Jan 13 at 22:08

add a comment |

Twig, 25 characters

{{d[:-3]|date('ymd.hi')}}

This is a template. Call it by including it and pass the Unix time as parameter d.

Sample usage:

{{include('datetime.twig', {'d': 1547233866744})}}

Try it on TwigFiddle

answered Jan 14 at 13:13

manatwork

16.3k43572

add a comment |

JavaScript, 64 bytes

n=>'2356891911121415'.replace(/1?./g,x=>new Date(n).toJSON()[x])

Try it online!

answered Jan 15 at 2:36

tsh

8,61511547

$begingroup$
aw, so close
$endgroup$
– ASCII-only
Jan 15 at 5:47

$begingroup$
also 64
$endgroup$
– ASCII-only
Jan 15 at 5:48

add a comment |

jq, 33 characters

(30 characters code + 3 characters command line option)

./1000|strftime("%y%m%d.%H%M")

Sample run:

bash-4.4$ jq -r './1000|strftime("%y%m%d.%H%M")' <<< 1547233866744

190111.1911

Try it online!

answered Jan 11 at 20:20

manatwork

16.3k43572

3

$begingroup$
You don't need to count command-line flags anymore.
$endgroup$
– AdmBorkBork
Jan 11 at 20:28

$begingroup$
Oops. Good to know. Thank you @AdmBorkBork.
$endgroup$
– manatwork
Jan 11 at 20:36

$begingroup$
1000 -> 1e3
$endgroup$
– Shaggy
Jan 14 at 20:16

add a comment |

ksh, 36 bytes

printf "%(%y%m%d.%H%M)T" $(($1/1e3))

Try it online!

Thanks to Jo King for 15 bytes saved

edited Jan 12 at 2:55

answered Jan 12 at 1:56

Sergiy Kolodyazhnyy

331110

$begingroup$
The same in Bash would be just 35 characters: Try it online!
$endgroup$
– manatwork
Jan 13 at 15:25

add a comment |

MediaWiki, 46 bytes

{{#time:ymd.Hi|@{{#expr:floor({{{1}}}/1e3)}}}}

answered Jan 13 at 11:19

tsh

8,61511547

add a comment |

Java 8, 78 bytes

n->new java.text.SimpleDateFormat("yyMMdd.HHmm").format(new java.util.Date(n))

Try it online.

Explanation:

n->                       // Method with long parameter and String return-type

  new java.text.SimpleDateFormat("yyMMdd.HHmm")

                          //  Create the formatter

   .format(               //  Format the date to a String in this format and return it:

     new java.util.Date(  //   Create a new Date

      n))                 //   With the input-long as timestamp

answered Jan 15 at 8:24

Kevin Cruijssen

36.8k555193

add a comment |

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21 Answers
21

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21 Answers
21

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oldest

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Japt v1.4.5, 19 16 bytes

GîÐU s3)¤o>J i.G

1 byte saved thanks to Oliver, which led to another 2 bytes saved.

Try it

Explanation

GîÐU s3)¤o>J i.G

                     :Implicit input of integer U

G                    :16

 î                   :Get the first 16 characters of the following string

  ÐU                 :  Convert U to a date object

     s3              :  Convert to ISO string

       )             :End get

        ¤            :Slice off the first 2 characters

         o           :Filter

          >J         :  Greater than -1

             i.G     :Insert "." at 0-based index 16, with wrapping

Notes / Tips

Getting the first 16 characters of the ISO string and then slicing off the first two is a byte shorter than performing a single slice.

G is used to insert the . at the required index because using a literal 6 would cause it to be combined with the . and for that to be interpreted as a decimal that would be inserted at the start of the string. To get around that I'd need to add a ' before the . to force it to be interpreted as a string.

Japt v1.4.5 is used because Japt doesn't have a constant for 17 and from v1.4.6 on trying to insert something at the first index past the end of a string results in it being inserted at the end of the string (A is the Japt constant for 10) whereas prior to v1.4.6 it immediately wraps back to the beginning of the string.

edited Jan 14 at 18:44

answered Jan 11 at 22:34

Shaggy

19.5k21666

add a comment |

Japt v1.4.5, 19 16 bytes

GîÐU s3)¤o>J i.G

1 byte saved thanks to Oliver, which led to another 2 bytes saved.

Try it

Explanation

GîÐU s3)¤o>J i.G

                     :Implicit input of integer U

G                    :16

 î                   :Get the first 16 characters of the following string

  ÐU                 :  Convert U to a date object

     s3              :  Convert to ISO string

       )             :End get

        ¤            :Slice off the first 2 characters

         o           :Filter

          >J         :  Greater than -1

             i.G     :Insert "." at 0-based index 16, with wrapping

Notes / Tips

Getting the first 16 characters of the ISO string and then slicing off the first two is a byte shorter than performing a single slice.

G is used to insert the . at the required index because using a literal 6 would cause it to be combined with the . and for that to be interpreted as a decimal that would be inserted at the start of the string. To get around that I'd need to add a ' before the . to force it to be interpreted as a string.

Japt v1.4.5 is used because Japt doesn't have a constant for 17 and from v1.4.6 on trying to insert something at the first index past the end of a string results in it being inserted at the end of the string (A is the Japt constant for 10) whereas prior to v1.4.6 it immediately wraps back to the beginning of the string.

edited Jan 14 at 18:44

answered Jan 11 at 22:34

Shaggy

19.5k21666

add a comment |

Japt v1.4.5, 19 16 bytes

GîÐU s3)¤o>J i.G

1 byte saved thanks to Oliver, which led to another 2 bytes saved.

Try it

Explanation

GîÐU s3)¤o>J i.G

                     :Implicit input of integer U

G                    :16

 î                   :Get the first 16 characters of the following string

  ÐU                 :  Convert U to a date object

     s3              :  Convert to ISO string

       )             :End get

        ¤            :Slice off the first 2 characters

         o           :Filter

          >J         :  Greater than -1

             i.G     :Insert "." at 0-based index 16, with wrapping

Notes / Tips

Getting the first 16 characters of the ISO string and then slicing off the first two is a byte shorter than performing a single slice.

G is used to insert the . at the required index because using a literal 6 would cause it to be combined with the . and for that to be interpreted as a decimal that would be inserted at the start of the string. To get around that I'd need to add a ' before the . to force it to be interpreted as a string.

Japt v1.4.5 is used because Japt doesn't have a constant for 17 and from v1.4.6 on trying to insert something at the first index past the end of a string results in it being inserted at the end of the string (A is the Japt constant for 10) whereas prior to v1.4.6 it immediately wraps back to the beginning of the string.

edited Jan 14 at 18:44

answered Jan 11 at 22:34

Shaggy

19.5k21666

Japt v1.4.5, 19 16 bytes

GîÐU s3)¤o>J i.G

1 byte saved thanks to Oliver, which led to another 2 bytes saved.

Try it

Explanation

GîÐU s3)¤o>J i.G

                     :Implicit input of integer U

G                    :16

 î                   :Get the first 16 characters of the following string

  ÐU                 :  Convert U to a date object

     s3              :  Convert to ISO string

       )             :End get

        ¤            :Slice off the first 2 characters

         o           :Filter

          >J         :  Greater than -1

             i.G     :Insert "." at 0-based index 16, with wrapping

Notes / Tips

Getting the first 16 characters of the ISO string and then slicing off the first two is a byte shorter than performing a single slice.

G is used to insert the . at the required index because using a literal 6 would cause it to be combined with the . and for that to be interpreted as a decimal that would be inserted at the start of the string. To get around that I'd need to add a ' before the . to force it to be interpreted as a string.

Japt v1.4.5 is used because Japt doesn't have a constant for 17 and from v1.4.6 on trying to insert something at the first index past the end of a string results in it being inserted at the end of the string (A is the Japt constant for 10) whereas prior to v1.4.6 it immediately wraps back to the beginning of the string.

edited Jan 14 at 18:44

answered Jan 11 at 22:34

Shaggy

19.5k21666

edited Jan 14 at 18:44

answered Jan 11 at 22:34

Shaggy

19.5k21666

answered Jan 11 at 22:34

Shaggy

19.5k21666

answered Jan 11 at 22:34

Shaggy

19.5k21666

add a comment |

JavaScript (ES6), 65 bytes

n=>'101010.1010'.replace(i=/d/g,x=>new Date(n).toJSON()[i=x-~i])

Try it online!

How?

We initialize the pointer $i$ to a non-numeric value (coerced to $0$) and then add alternately $2$ and $1$ to it to pick the relevant characters from the ISO-8601 conversion of the input timestamp.

yyyy-mm-ddThh:mm:ss.sssZ

  ^^ ^^ ^^ ^^ ^^

JavaScript (ES6), 66 bytes

n=>'235689.BCEF'.replace(/w/g,x=>new Date(n).toJSON()[+`0x${x}`])

Try it online!

How?

Once the input timestamp is converted in ISO-8601 format, all required characters can be accessed with a single hexadecimal digit.

yyyy-mm-ddThh:mm:ss.sssZ

  ↓↓ ↓↓ ↓↓ ↓↓ ↓↓

0123456789ABCDEF

edited Jan 12 at 20:49

answered Jan 12 at 10:51

Arnauld

74.1k690310

$begingroup$
Wow. Just, wow.
$endgroup$
– skiilaa
Jan 13 at 9:08

1

$begingroup$
64 bytes: n=>'235689.11121415'.replace(/1?w/g,x=>new Date(n).toJSON()[x])
$endgroup$
– tsh
Jan 14 at 7:02

$begingroup$
@tsh Nice! I was not expecting the decimal format to be shorter. You may want to post this as a separate answer.
$endgroup$
– Arnauld
Jan 14 at 15:45

add a comment |

JavaScript (ES6), 65 bytes

n=>'101010.1010'.replace(i=/d/g,x=>new Date(n).toJSON()[i=x-~i])

Try it online!

How?

We initialize the pointer $i$ to a non-numeric value (coerced to $0$) and then add alternately $2$ and $1$ to it to pick the relevant characters from the ISO-8601 conversion of the input timestamp.

yyyy-mm-ddThh:mm:ss.sssZ

  ^^ ^^ ^^ ^^ ^^

JavaScript (ES6), 66 bytes

n=>'235689.BCEF'.replace(/w/g,x=>new Date(n).toJSON()[+`0x${x}`])

Try it online!

How?

Once the input timestamp is converted in ISO-8601 format, all required characters can be accessed with a single hexadecimal digit.

yyyy-mm-ddThh:mm:ss.sssZ

  ↓↓ ↓↓ ↓↓ ↓↓ ↓↓

0123456789ABCDEF

edited Jan 12 at 20:49

answered Jan 12 at 10:51

Arnauld

74.1k690310

$begingroup$
Wow. Just, wow.
$endgroup$
– skiilaa
Jan 13 at 9:08

1

$begingroup$
64 bytes: n=>'235689.11121415'.replace(/1?w/g,x=>new Date(n).toJSON()[x])
$endgroup$
– tsh
Jan 14 at 7:02

$begingroup$
@tsh Nice! I was not expecting the decimal format to be shorter. You may want to post this as a separate answer.
$endgroup$
– Arnauld
Jan 14 at 15:45

add a comment |

JavaScript (ES6), 65 bytes

n=>'101010.1010'.replace(i=/d/g,x=>new Date(n).toJSON()[i=x-~i])

Try it online!

How?

We initialize the pointer $i$ to a non-numeric value (coerced to $0$) and then add alternately $2$ and $1$ to it to pick the relevant characters from the ISO-8601 conversion of the input timestamp.

yyyy-mm-ddThh:mm:ss.sssZ

  ^^ ^^ ^^ ^^ ^^

JavaScript (ES6), 66 bytes

n=>'235689.BCEF'.replace(/w/g,x=>new Date(n).toJSON()[+`0x${x}`])

Try it online!

How?

Once the input timestamp is converted in ISO-8601 format, all required characters can be accessed with a single hexadecimal digit.

yyyy-mm-ddThh:mm:ss.sssZ

  ↓↓ ↓↓ ↓↓ ↓↓ ↓↓

0123456789ABCDEF

edited Jan 12 at 20:49

answered Jan 12 at 10:51

Arnauld

74.1k690310

JavaScript (ES6), 65 bytes

n=>'101010.1010'.replace(i=/d/g,x=>new Date(n).toJSON()[i=x-~i])

Try it online!

How?

We initialize the pointer $i$ to a non-numeric value (coerced to $0$) and then add alternately $2$ and $1$ to it to pick the relevant characters from the ISO-8601 conversion of the input timestamp.

yyyy-mm-ddThh:mm:ss.sssZ

  ^^ ^^ ^^ ^^ ^^

JavaScript (ES6), 66 bytes

n=>'235689.BCEF'.replace(/w/g,x=>new Date(n).toJSON()[+`0x${x}`])

Try it online!

How?

Once the input timestamp is converted in ISO-8601 format, all required characters can be accessed with a single hexadecimal digit.

yyyy-mm-ddThh:mm:ss.sssZ

  ↓↓ ↓↓ ↓↓ ↓↓ ↓↓

0123456789ABCDEF

edited Jan 12 at 20:49

answered Jan 12 at 10:51

Arnauld

74.1k690310

edited Jan 12 at 20:49

answered Jan 12 at 10:51

Arnauld

74.1k690310

answered Jan 12 at 10:51

Arnauld

74.1k690310

answered Jan 12 at 10:51

Arnauld

74.1k690310

$begingroup$
Wow. Just, wow.
$endgroup$
– skiilaa
Jan 13 at 9:08

1

$begingroup$
64 bytes: n=>'235689.11121415'.replace(/1?w/g,x=>new Date(n).toJSON()[x])
$endgroup$
– tsh
Jan 14 at 7:02

$begingroup$
@tsh Nice! I was not expecting the decimal format to be shorter. You may want to post this as a separate answer.
$endgroup$
– Arnauld
Jan 14 at 15:45

add a comment |

$begingroup$
Wow. Just, wow.
$endgroup$
– skiilaa
Jan 13 at 9:08

1

$begingroup$
64 bytes: n=>'235689.11121415'.replace(/1?w/g,x=>new Date(n).toJSON()[x])
$endgroup$
– tsh
Jan 14 at 7:02

$begingroup$
@tsh Nice! I was not expecting the decimal format to be shorter. You may want to post this as a separate answer.
$endgroup$
– Arnauld
Jan 14 at 15:45

Wow. Just, wow.

– skiilaa
Jan 13 at 9:08

64 bytes: n=>'235689.11121415'.replace(/1?w/g,x=>new Date(n).toJSON()[x])

– tsh
Jan 14 at 7:02

@tsh Nice! I was not expecting the decimal format to be shorter. You may want to post this as a separate answer.

– Arnauld
Jan 14 at 15:45

add a comment |

Bash + coreutils, 29 bytes

date -d@${1::-3} +%y%m%d.%H%M

Try it online!

answered Jan 11 at 19:59

Neil

80k744178

add a comment |

Bash + coreutils, 29 bytes

date -d@${1::-3} +%y%m%d.%H%M

Try it online!

answered Jan 11 at 19:59

Neil

80k744178

add a comment |

Bash + coreutils, 29 bytes

date -d@${1::-3} +%y%m%d.%H%M

Try it online!

answered Jan 11 at 19:59

Neil

80k744178

Bash + coreutils, 29 bytes

date -d@${1::-3} +%y%m%d.%H%M

Try it online!

answered Jan 11 at 19:59

Neil

80k744178

answered Jan 11 at 19:59

Neil

80k744178

answered Jan 11 at 19:59

Neil

80k744178

answered Jan 11 at 19:59

Neil

80k744178

add a comment |

PHP, 40 32 31 bytes

-8 bytes thanks to Luis felipe
-1 byte thanks to Jo King

<?=date('ymd.hi',$argv[1]/1e3);

Try it online!

edited Jan 14 at 14:02

answered Jan 11 at 20:00

Skidsdev

6,4262974

$begingroup$
Nice answer! It's a tad shorter if the language has date formatting, yes :)
$endgroup$
– skiilaa
Jan 11 at 20:03

$begingroup$
32 bytes you dont need to cast to integer since "2"/1 will cast automatically the string
$endgroup$
– Luis felipe De jesus Munoz
Jan 11 at 20:04

$begingroup$
ah of course, I added intval before realising I needed /1000, and didn't think that I mihgt not need it afterwards :P
$endgroup$
– Skidsdev
Jan 11 at 20:08

$begingroup$
@JoKing ah nice one, thanks
$endgroup$
– Skidsdev
Jan 14 at 14:03

add a comment |

PHP, 40 32 31 bytes

-8 bytes thanks to Luis felipe
-1 byte thanks to Jo King

<?=date('ymd.hi',$argv[1]/1e3);

Try it online!

edited Jan 14 at 14:02

answered Jan 11 at 20:00

Skidsdev

6,4262974

$begingroup$
Nice answer! It's a tad shorter if the language has date formatting, yes :)
$endgroup$
– skiilaa
Jan 11 at 20:03

$begingroup$
32 bytes you dont need to cast to integer since "2"/1 will cast automatically the string
$endgroup$
– Luis felipe De jesus Munoz
Jan 11 at 20:04

$begingroup$
ah of course, I added intval before realising I needed /1000, and didn't think that I mihgt not need it afterwards :P
$endgroup$
– Skidsdev
Jan 11 at 20:08

$begingroup$
@JoKing ah nice one, thanks
$endgroup$
– Skidsdev
Jan 14 at 14:03

add a comment |

PHP, 40 32 31 bytes

-8 bytes thanks to Luis felipe
-1 byte thanks to Jo King

<?=date('ymd.hi',$argv[1]/1e3);

Try it online!

edited Jan 14 at 14:02

answered Jan 11 at 20:00

Skidsdev

6,4262974

PHP, 40 32 31 bytes

-8 bytes thanks to Luis felipe
-1 byte thanks to Jo King

<?=date('ymd.hi',$argv[1]/1e3);

Try it online!

edited Jan 14 at 14:02

answered Jan 11 at 20:00

Skidsdev

6,4262974

edited Jan 14 at 14:02

answered Jan 11 at 20:00

Skidsdev

6,4262974

answered Jan 11 at 20:00

Skidsdev

6,4262974

answered Jan 11 at 20:00

Skidsdev

6,4262974

$begingroup$
Nice answer! It's a tad shorter if the language has date formatting, yes :)
$endgroup$
– skiilaa
Jan 11 at 20:03

$begingroup$
32 bytes you dont need to cast to integer since "2"/1 will cast automatically the string
$endgroup$
– Luis felipe De jesus Munoz
Jan 11 at 20:04

$begingroup$
ah of course, I added intval before realising I needed /1000, and didn't think that I mihgt not need it afterwards :P
$endgroup$
– Skidsdev
Jan 11 at 20:08

$begingroup$
@JoKing ah nice one, thanks
$endgroup$
– Skidsdev
Jan 14 at 14:03

add a comment |

$begingroup$
Nice answer! It's a tad shorter if the language has date formatting, yes :)
$endgroup$
– skiilaa
Jan 11 at 20:03

$begingroup$
32 bytes you dont need to cast to integer since "2"/1 will cast automatically the string
$endgroup$
– Luis felipe De jesus Munoz
Jan 11 at 20:04

$begingroup$
ah of course, I added intval before realising I needed /1000, and didn't think that I mihgt not need it afterwards :P
$endgroup$
– Skidsdev
Jan 11 at 20:08

$begingroup$
@JoKing ah nice one, thanks
$endgroup$
– Skidsdev
Jan 14 at 14:03

Nice answer! It's a tad shorter if the language has date formatting, yes :)

– skiilaa
Jan 11 at 20:03

32 bytes you dont need to cast to integer since "2"/1 will cast automatically the string

– Luis felipe De jesus Munoz
Jan 11 at 20:04

ah of course, I added intval before realising I needed /1000, and didn't think that I mihgt not need it afterwards :P

– Skidsdev
Jan 11 at 20:08

@JoKing ah nice one, thanks

– Skidsdev
Jan 14 at 14:03

add a comment |

MATL, 28 bytes

Thanks to @skiilaa for a correction in the output format.

864e5/719529+'YYmmDD.HHMM'XO

Try it online!

Explanation

MATL, like MATLAB, defines date/time numbers as the (possibly non-integer) number of days since time 00:00 of the reference "date" 0-Jan-0000.

edited Jan 13 at 16:06

answered Jan 11 at 22:00

Luis Mendo

74.2k887291

add a comment |

MATL, 28 bytes

Thanks to @skiilaa for a correction in the output format.

864e5/719529+'YYmmDD.HHMM'XO

Try it online!

Explanation

MATL, like MATLAB, defines date/time numbers as the (possibly non-integer) number of days since time 00:00 of the reference "date" 0-Jan-0000.

edited Jan 13 at 16:06

answered Jan 11 at 22:00

Luis Mendo

74.2k887291

add a comment |

MATL, 28 bytes

Thanks to @skiilaa for a correction in the output format.

864e5/719529+'YYmmDD.HHMM'XO

Try it online!

Explanation

MATL, like MATLAB, defines date/time numbers as the (possibly non-integer) number of days since time 00:00 of the reference "date" 0-Jan-0000.

edited Jan 13 at 16:06

answered Jan 11 at 22:00

Luis Mendo

74.2k887291

MATL, 28 bytes

Thanks to @skiilaa for a correction in the output format.

864e5/719529+'YYmmDD.HHMM'XO

Try it online!

Explanation

MATL, like MATLAB, defines date/time numbers as the (possibly non-integer) number of days since time 00:00 of the reference "date" 0-Jan-0000.

edited Jan 13 at 16:06

answered Jan 11 at 22:00

Luis Mendo

74.2k887291

edited Jan 13 at 16:06

answered Jan 11 at 22:00

Luis Mendo

74.2k887291

answered Jan 11 at 22:00

Luis Mendo

74.2k887291

answered Jan 11 at 22:00

Luis Mendo

74.2k887291

add a comment |

GNU AWK, 34 33 characters

$0=strftime("%y%m%d.%H%M",$0/1e3)

(strftime() is GNU extension, will not run in other AWK implementations.)

Thanks to:

Jo King for suggesting E-notation (-1 character)

Sampler run:

bash-4.4$ awk '$0=strftime("%y%m%d.%H%M",$0/1e3)' <<< 1547233866744

190111.2111

Try it online!

edited Jan 13 at 16:05

answered Jan 11 at 20:28

manatwork

16.3k43572

$begingroup$
Thank you, @JoKing. Unfortunately E-notation is not really my friend so I always forget it.
$endgroup$
– manatwork
Jan 13 at 16:02

add a comment |

GNU AWK, 34 33 characters

$0=strftime("%y%m%d.%H%M",$0/1e3)

(strftime() is GNU extension, will not run in other AWK implementations.)

Thanks to:

Jo King for suggesting E-notation (-1 character)

Sampler run:

bash-4.4$ awk '$0=strftime("%y%m%d.%H%M",$0/1e3)' <<< 1547233866744

190111.2111

Try it online!

edited Jan 13 at 16:05

answered Jan 11 at 20:28

manatwork

16.3k43572

$begingroup$
Thank you, @JoKing. Unfortunately E-notation is not really my friend so I always forget it.
$endgroup$
– manatwork
Jan 13 at 16:02

add a comment |

GNU AWK, 34 33 characters

$0=strftime("%y%m%d.%H%M",$0/1e3)

(strftime() is GNU extension, will not run in other AWK implementations.)

Thanks to:

Jo King for suggesting E-notation (-1 character)

Sampler run:

bash-4.4$ awk '$0=strftime("%y%m%d.%H%M",$0/1e3)' <<< 1547233866744

190111.2111

Try it online!

edited Jan 13 at 16:05

answered Jan 11 at 20:28

manatwork

16.3k43572

GNU AWK, 34 33 characters

$0=strftime("%y%m%d.%H%M",$0/1e3)

(strftime() is GNU extension, will not run in other AWK implementations.)

Thanks to:

Jo King for suggesting E-notation (-1 character)

Sampler run:

bash-4.4$ awk '$0=strftime("%y%m%d.%H%M",$0/1e3)' <<< 1547233866744

190111.2111

Try it online!

edited Jan 13 at 16:05

answered Jan 11 at 20:28

manatwork

16.3k43572

edited Jan 13 at 16:05

answered Jan 11 at 20:28

manatwork

16.3k43572

answered Jan 11 at 20:28

manatwork

16.3k43572

answered Jan 11 at 20:28

manatwork

16.3k43572

$begingroup$
Thank you, @JoKing. Unfortunately E-notation is not really my friend so I always forget it.
$endgroup$
– manatwork
Jan 13 at 16:02

add a comment |

$begingroup$
Thank you, @JoKing. Unfortunately E-notation is not really my friend so I always forget it.
$endgroup$
– manatwork
Jan 13 at 16:02

Thank you, @JoKing. Unfortunately E-notation is not really my friend so I always forget it.

– manatwork
Jan 13 at 16:02

add a comment |

PowerShell, 59 58 bytes

"{0:yyMMdd.HHmm}"-f(Date 1/1/1970).AddSeconds("$args"/1e3)

Try it online!

Gets the Date of 1/1/1970 (defaults to 00:00:00am), then Adds the appropriate number of Seconds. Passes that to the -format operator, which correctly formats the datetime.

Probably culture-dependent. This works on TIO, which is en-us.

-1 byte thanks to shaggy.

edited Jan 14 at 13:39

answered Jan 11 at 20:15

AdmBorkBork

26.6k364229

add a comment |

PowerShell, 59 58 bytes

"{0:yyMMdd.HHmm}"-f(Date 1/1/1970).AddSeconds("$args"/1e3)

Try it online!

Gets the Date of 1/1/1970 (defaults to 00:00:00am), then Adds the appropriate number of Seconds. Passes that to the -format operator, which correctly formats the datetime.

Probably culture-dependent. This works on TIO, which is en-us.

-1 byte thanks to shaggy.

edited Jan 14 at 13:39

answered Jan 11 at 20:15

AdmBorkBork

26.6k364229

add a comment |

PowerShell, 59 58 bytes

"{0:yyMMdd.HHmm}"-f(Date 1/1/1970).AddSeconds("$args"/1e3)

Try it online!

Gets the Date of 1/1/1970 (defaults to 00:00:00am), then Adds the appropriate number of Seconds. Passes that to the -format operator, which correctly formats the datetime.

Probably culture-dependent. This works on TIO, which is en-us.

-1 byte thanks to shaggy.

edited Jan 14 at 13:39

answered Jan 11 at 20:15

AdmBorkBork

26.6k364229

PowerShell, 59 58 bytes

"{0:yyMMdd.HHmm}"-f(Date 1/1/1970).AddSeconds("$args"/1e3)

Try it online!

Gets the Date of 1/1/1970 (defaults to 00:00:00am), then Adds the appropriate number of Seconds. Passes that to the -format operator, which correctly formats the datetime.

Probably culture-dependent. This works on TIO, which is en-us.

-1 byte thanks to shaggy.

edited Jan 14 at 13:39

answered Jan 11 at 20:15

AdmBorkBork

26.6k364229

edited Jan 14 at 13:39

answered Jan 11 at 20:15

AdmBorkBork

26.6k364229

answered Jan 11 at 20:15

AdmBorkBork

26.6k364229

answered Jan 11 at 20:15

AdmBorkBork

26.6k364229

add a comment |

Perl 6, 111 89 87 bytes

{~DateTime.new($_/Ⅿ,:formatter{"{(.year%Ⅽ,.month,.day).fmt('%02d','')}.{(.hour,.minute).fmt('%02d','')}"})}

Try it (111)

{TR/-//}o{S/..//}o{.yyyy-mm-dd~'.'~(.hour,.minute).fmt('%02d','')}o{DateTime.new($_/Ⅿ)}

Try it (89)

{TR/- //}o{S/..//}o{.yyyy-mm-dd~'.'~(.hour,.minute).fmt('%02d')}o{DateTime.new($_/Ⅿ)}

Try it (87)

Explanation:

The o infix operator takes two functions and creates a composite function. The rightmost one gets called first, and the one to the left gets called with its result.

Basically we use 4 block lambdas to generate a single lambda.

Which is not much different to how a WhateverCode lambda like * + * gets created.

Divide by 1000 and use that to create a DateTime object.

{DateTime.new($_/Ⅿ)} # Ⅿ is ROMAN NUMERAL ONE THOUSAND (3 bytes)

The result gets used by:

{

   .yyyy-mm-dd # 2019-01-11



   ~ '.' ~     # str concatenation with '.'



   ( .hour, .minute ).fmt('%02d') # add leading 0s (returns List)

}

That leaves us with a string like 2019-01-11.19 11

We need to remove the first two digits

{S/..//}

We also need to remove - and

{TR/- //}

edited Jan 11 at 21:37

answered Jan 11 at 20:46

Brad Gilbert b2gills

12.2k11232

add a comment |

Perl 6, 111 89 87 bytes

{~DateTime.new($_/Ⅿ,:formatter{"{(.year%Ⅽ,.month,.day).fmt('%02d','')}.{(.hour,.minute).fmt('%02d','')}"})}

Try it (111)

{TR/-//}o{S/..//}o{.yyyy-mm-dd~'.'~(.hour,.minute).fmt('%02d','')}o{DateTime.new($_/Ⅿ)}

Try it (89)

{TR/- //}o{S/..//}o{.yyyy-mm-dd~'.'~(.hour,.minute).fmt('%02d')}o{DateTime.new($_/Ⅿ)}

Try it (87)

Explanation:

The o infix operator takes two functions and creates a composite function. The rightmost one gets called first, and the one to the left gets called with its result.

Basically we use 4 block lambdas to generate a single lambda.

Which is not much different to how a WhateverCode lambda like * + * gets created.

Divide by 1000 and use that to create a DateTime object.

{DateTime.new($_/Ⅿ)} # Ⅿ is ROMAN NUMERAL ONE THOUSAND (3 bytes)

The result gets used by:

{

   .yyyy-mm-dd # 2019-01-11



   ~ '.' ~     # str concatenation with '.'



   ( .hour, .minute ).fmt('%02d') # add leading 0s (returns List)

}

That leaves us with a string like 2019-01-11.19 11

We need to remove the first two digits

{S/..//}

We also need to remove - and

{TR/- //}

edited Jan 11 at 21:37

answered Jan 11 at 20:46

Brad Gilbert b2gills

12.2k11232

add a comment |

Perl 6, 111 89 87 bytes

{~DateTime.new($_/Ⅿ,:formatter{"{(.year%Ⅽ,.month,.day).fmt('%02d','')}.{(.hour,.minute).fmt('%02d','')}"})}

Try it (111)

{TR/-//}o{S/..//}o{.yyyy-mm-dd~'.'~(.hour,.minute).fmt('%02d','')}o{DateTime.new($_/Ⅿ)}

Try it (89)

{TR/- //}o{S/..//}o{.yyyy-mm-dd~'.'~(.hour,.minute).fmt('%02d')}o{DateTime.new($_/Ⅿ)}

Try it (87)

Explanation:

The o infix operator takes two functions and creates a composite function. The rightmost one gets called first, and the one to the left gets called with its result.

Basically we use 4 block lambdas to generate a single lambda.

Which is not much different to how a WhateverCode lambda like * + * gets created.

Divide by 1000 and use that to create a DateTime object.

{DateTime.new($_/Ⅿ)} # Ⅿ is ROMAN NUMERAL ONE THOUSAND (3 bytes)

The result gets used by:

{

   .yyyy-mm-dd # 2019-01-11



   ~ '.' ~     # str concatenation with '.'



   ( .hour, .minute ).fmt('%02d') # add leading 0s (returns List)

}

That leaves us with a string like 2019-01-11.19 11

We need to remove the first two digits

{S/..//}

We also need to remove - and

{TR/- //}

edited Jan 11 at 21:37

answered Jan 11 at 20:46

Brad Gilbert b2gills

12.2k11232

Perl 6, 111 89 87 bytes

{~DateTime.new($_/Ⅿ,:formatter{"{(.year%Ⅽ,.month,.day).fmt('%02d','')}.{(.hour,.minute).fmt('%02d','')}"})}

Try it (111)

{TR/-//}o{S/..//}o{.yyyy-mm-dd~'.'~(.hour,.minute).fmt('%02d','')}o{DateTime.new($_/Ⅿ)}

Try it (89)

{TR/- //}o{S/..//}o{.yyyy-mm-dd~'.'~(.hour,.minute).fmt('%02d')}o{DateTime.new($_/Ⅿ)}

Try it (87)

Explanation:

The o infix operator takes two functions and creates a composite function. The rightmost one gets called first, and the one to the left gets called with its result.

Basically we use 4 block lambdas to generate a single lambda.

Which is not much different to how a WhateverCode lambda like * + * gets created.

Divide by 1000 and use that to create a DateTime object.

{DateTime.new($_/Ⅿ)} # Ⅿ is ROMAN NUMERAL ONE THOUSAND (3 bytes)

The result gets used by:

{

   .yyyy-mm-dd # 2019-01-11



   ~ '.' ~     # str concatenation with '.'



   ( .hour, .minute ).fmt('%02d') # add leading 0s (returns List)

}

That leaves us with a string like 2019-01-11.19 11

We need to remove the first two digits

{S/..//}

We also need to remove - and

{TR/- //}

edited Jan 11 at 21:37

answered Jan 11 at 20:46

Brad Gilbert b2gills

12.2k11232

edited Jan 11 at 21:37

answered Jan 11 at 20:46

Brad Gilbert b2gills

12.2k11232

answered Jan 11 at 20:46

Brad Gilbert b2gills

12.2k11232

answered Jan 11 at 20:46

Brad Gilbert b2gills

12.2k11232

add a comment |

Python 2, 64 bytes

lambda s:strftime('%y%m%d.%H%M',gmtime(s/1e3))

from time import*

Try it online!

The input is considered to be in UTC.

edited Jan 11 at 22:01

answered Jan 11 at 21:55

Erik the Outgolfer

31.6k429103

1

$begingroup$
Just as something I've been a bit confused on, is the only reason that the lambda comes before the import because it works better for TIO's header?
$endgroup$
– Neil A.
Jan 11 at 22:43

2

$begingroup$
@NeilA. Yes. The import can go either before or after, the contents of the lambda aren't checked for NameErrors before it's called.
$endgroup$
– Erik the Outgolfer
Jan 11 at 22:45

add a comment |

Python 2, 64 bytes

lambda s:strftime('%y%m%d.%H%M',gmtime(s/1e3))

from time import*

Try it online!

The input is considered to be in UTC.

edited Jan 11 at 22:01

answered Jan 11 at 21:55

Erik the Outgolfer

31.6k429103

1

$begingroup$
Just as something I've been a bit confused on, is the only reason that the lambda comes before the import because it works better for TIO's header?
$endgroup$
– Neil A.
Jan 11 at 22:43

2

$begingroup$
@NeilA. Yes. The import can go either before or after, the contents of the lambda aren't checked for NameErrors before it's called.
$endgroup$
– Erik the Outgolfer
Jan 11 at 22:45

add a comment |

Python 2, 64 bytes

lambda s:strftime('%y%m%d.%H%M',gmtime(s/1e3))

from time import*

Try it online!

The input is considered to be in UTC.

edited Jan 11 at 22:01

answered Jan 11 at 21:55

Erik the Outgolfer

31.6k429103

Python 2, 64 bytes

lambda s:strftime('%y%m%d.%H%M',gmtime(s/1e3))

from time import*

Try it online!

The input is considered to be in UTC.

edited Jan 11 at 22:01

answered Jan 11 at 21:55

Erik the Outgolfer

31.6k429103

edited Jan 11 at 22:01

answered Jan 11 at 21:55

Erik the Outgolfer

31.6k429103

answered Jan 11 at 21:55

Erik the Outgolfer

31.6k429103

answered Jan 11 at 21:55

Erik the Outgolfer

31.6k429103

1

$begingroup$
Just as something I've been a bit confused on, is the only reason that the lambda comes before the import because it works better for TIO's header?
$endgroup$
– Neil A.
Jan 11 at 22:43

2

$begingroup$
@NeilA. Yes. The import can go either before or after, the contents of the lambda aren't checked for NameErrors before it's called.
$endgroup$
– Erik the Outgolfer
Jan 11 at 22:45

add a comment |

1

$begingroup$
Just as something I've been a bit confused on, is the only reason that the lambda comes before the import because it works better for TIO's header?
$endgroup$
– Neil A.
Jan 11 at 22:43

2

$begingroup$
@NeilA. Yes. The import can go either before or after, the contents of the lambda aren't checked for NameErrors before it's called.
$endgroup$
– Erik the Outgolfer
Jan 11 at 22:45

Just as something I've been a bit confused on, is the only reason that the lambda comes before the import because it works better for TIO's header?

– Neil A.
Jan 11 at 22:43

@NeilA. Yes. The import can go either before or after, the contents of the lambda aren't checked for NameErrors before it's called.

– Erik the Outgolfer
Jan 11 at 22:45

add a comment |

R, 58 56 bytes

format(as.POSIXct(scan()/1e3,,'1970-1-1'),'%y%m%d.%H%M')

Try it online!

edited Jan 12 at 12:03

answered Jan 12 at 11:22

digEmAll

3,029414

add a comment |

R, 58 56 bytes

format(as.POSIXct(scan()/1e3,,'1970-1-1'),'%y%m%d.%H%M')

Try it online!

edited Jan 12 at 12:03

answered Jan 12 at 11:22

digEmAll

3,029414

add a comment |

R, 58 56 bytes

format(as.POSIXct(scan()/1e3,,'1970-1-1'),'%y%m%d.%H%M')

Try it online!

edited Jan 12 at 12:03

answered Jan 12 at 11:22

digEmAll

3,029414

R, 58 56 bytes

format(as.POSIXct(scan()/1e3,,'1970-1-1'),'%y%m%d.%H%M')

Try it online!

edited Jan 12 at 12:03

answered Jan 12 at 11:22

digEmAll

3,029414

edited Jan 12 at 12:03

answered Jan 12 at 11:22

digEmAll

3,029414

answered Jan 12 at 11:22

digEmAll

3,029414

answered Jan 12 at 11:22

digEmAll

3,029414

add a comment |

C (gcc) (32-bit, little endian), 67 bytes

f(t,s)long long t;{t/=1e3;strftime(s,12,"%y%m%d.%H%M",gmtime(&t));}

Try it online!

On an ILP64 platform, the following 55 byte version should work:

f(t,s){t/=1e3;strftime(s,12,"%y%m%d.%H%M",gmtime(&t));}

edited Jan 12 at 15:08

answered Jan 12 at 14:32

nwellnhof

6,67511126

$begingroup$
What's the extra s argument you're taking for?
$endgroup$
– Shaggy
Jan 12 at 14:47

1

$begingroup$
@Shaggy The s is for the output string.
$endgroup$
– nwellnhof
Jan 12 at 15:06

$begingroup$
It looks like you're initiating it outside the function; do we have a meta consensus that allows C to do that?
$endgroup$
– Shaggy
Jan 12 at 15:31

$begingroup$
@Shaggy I'm not sure what "consensus" means exactly but here's the relevant meta post.
$endgroup$
– nwellnhof
Jan 12 at 15:54

$begingroup$
With "32-bit,", why long long instead of long or int32_t, int64_t, time_t? IAC, all shorter than long long.
$endgroup$
– chux
Jan 13 at 22:23

|
show 2 more comments

C (gcc) (32-bit, little endian), 67 bytes

f(t,s)long long t;{t/=1e3;strftime(s,12,"%y%m%d.%H%M",gmtime(&t));}

Try it online!

On an ILP64 platform, the following 55 byte version should work:

f(t,s){t/=1e3;strftime(s,12,"%y%m%d.%H%M",gmtime(&t));}

edited Jan 12 at 15:08

answered Jan 12 at 14:32

nwellnhof

6,67511126

$begingroup$
What's the extra s argument you're taking for?
$endgroup$
– Shaggy
Jan 12 at 14:47

1

$begingroup$
@Shaggy The s is for the output string.
$endgroup$
– nwellnhof
Jan 12 at 15:06

$begingroup$
It looks like you're initiating it outside the function; do we have a meta consensus that allows C to do that?
$endgroup$
– Shaggy
Jan 12 at 15:31

$begingroup$
@Shaggy I'm not sure what "consensus" means exactly but here's the relevant meta post.
$endgroup$
– nwellnhof
Jan 12 at 15:54

$begingroup$
With "32-bit,", why long long instead of long or int32_t, int64_t, time_t? IAC, all shorter than long long.
$endgroup$
– chux
Jan 13 at 22:23

|
show 2 more comments

C (gcc) (32-bit, little endian), 67 bytes

f(t,s)long long t;{t/=1e3;strftime(s,12,"%y%m%d.%H%M",gmtime(&t));}

Try it online!

On an ILP64 platform, the following 55 byte version should work:

f(t,s){t/=1e3;strftime(s,12,"%y%m%d.%H%M",gmtime(&t));}

edited Jan 12 at 15:08

answered Jan 12 at 14:32

nwellnhof

6,67511126

C (gcc) (32-bit, little endian), 67 bytes

f(t,s)long long t;{t/=1e3;strftime(s,12,"%y%m%d.%H%M",gmtime(&t));}

Try it online!

On an ILP64 platform, the following 55 byte version should work:

f(t,s){t/=1e3;strftime(s,12,"%y%m%d.%H%M",gmtime(&t));}

edited Jan 12 at 15:08

answered Jan 12 at 14:32

nwellnhof

6,67511126

edited Jan 12 at 15:08

answered Jan 12 at 14:32

nwellnhof

6,67511126

answered Jan 12 at 14:32

nwellnhof

6,67511126

answered Jan 12 at 14:32

nwellnhof

6,67511126

$begingroup$
What's the extra s argument you're taking for?
$endgroup$
– Shaggy
Jan 12 at 14:47

1

$begingroup$
@Shaggy The s is for the output string.
$endgroup$
– nwellnhof
Jan 12 at 15:06

$begingroup$
It looks like you're initiating it outside the function; do we have a meta consensus that allows C to do that?
$endgroup$
– Shaggy
Jan 12 at 15:31

$begingroup$
@Shaggy I'm not sure what "consensus" means exactly but here's the relevant meta post.
$endgroup$
– nwellnhof
Jan 12 at 15:54

$begingroup$
With "32-bit,", why long long instead of long or int32_t, int64_t, time_t? IAC, all shorter than long long.
$endgroup$
– chux
Jan 13 at 22:23

|
show 2 more comments

$begingroup$
What's the extra s argument you're taking for?
$endgroup$
– Shaggy
Jan 12 at 14:47

1

$begingroup$
@Shaggy The s is for the output string.
$endgroup$
– nwellnhof
Jan 12 at 15:06

$begingroup$
It looks like you're initiating it outside the function; do we have a meta consensus that allows C to do that?
$endgroup$
– Shaggy
Jan 12 at 15:31

$begingroup$
@Shaggy I'm not sure what "consensus" means exactly but here's the relevant meta post.
$endgroup$
– nwellnhof
Jan 12 at 15:54

$begingroup$
With "32-bit,", why long long instead of long or int32_t, int64_t, time_t? IAC, all shorter than long long.
$endgroup$
– chux
Jan 13 at 22:23

What's the extra s argument you're taking for?

– Shaggy
Jan 12 at 14:47

@Shaggy The s is for the output string.

– nwellnhof
Jan 12 at 15:06

It looks like you're initiating it outside the function; do we have a meta consensus that allows C to do that?

– Shaggy
Jan 12 at 15:31

@Shaggy I'm not sure what "consensus" means exactly but here's the relevant meta post.

– nwellnhof
Jan 12 at 15:54

With "32-bit,", why long long instead of long or int32_t, int64_t, time_t? IAC, all shorter than long long.

– chux
Jan 13 at 22:23

|
show 2 more comments

Perl 6, 57 50 bytes

{TR:d/T:-/./}o{substr ~DateTime.new($_/1e3): 2,15}

Try it online!

Takes the default stringification of a Datetime, in the format yyyy-mm-ddThh:mm:ssZ and modifies it to fit the output format. Perl 6 is in need of a date formatter method.

Explanation:

                       Datetime.new($_/1e3) # Create a date time

                      ~                 # Stringify it to the format yyyy-mm-ddThh:mm:ssZ

                                        # e.g. 2019-01-11T19:11:06.744000Z

               substr                      : 2,15  # Take the middle 15 characters

 {TR:d/T  /./}o   # Then replace 'T' with '.'

        :-        # Then remove ':' and '-'

edited Jan 11 at 23:38

answered Jan 11 at 22:50

Jo King

21.5k248110

add a comment |

Perl 6, 57 50 bytes

{TR:d/T:-/./}o{substr ~DateTime.new($_/1e3): 2,15}

Try it online!

Takes the default stringification of a Datetime, in the format yyyy-mm-ddThh:mm:ssZ and modifies it to fit the output format. Perl 6 is in need of a date formatter method.

Explanation:

                       Datetime.new($_/1e3) # Create a date time

                      ~                 # Stringify it to the format yyyy-mm-ddThh:mm:ssZ

                                        # e.g. 2019-01-11T19:11:06.744000Z

               substr                      : 2,15  # Take the middle 15 characters

 {TR:d/T  /./}o   # Then replace 'T' with '.'

        :-        # Then remove ':' and '-'

edited Jan 11 at 23:38

answered Jan 11 at 22:50

Jo King

21.5k248110

add a comment |

Perl 6, 57 50 bytes

{TR:d/T:-/./}o{substr ~DateTime.new($_/1e3): 2,15}

Try it online!

Takes the default stringification of a Datetime, in the format yyyy-mm-ddThh:mm:ssZ and modifies it to fit the output format. Perl 6 is in need of a date formatter method.

Explanation:

                       Datetime.new($_/1e3) # Create a date time

                      ~                 # Stringify it to the format yyyy-mm-ddThh:mm:ssZ

                                        # e.g. 2019-01-11T19:11:06.744000Z

               substr                      : 2,15  # Take the middle 15 characters

 {TR:d/T  /./}o   # Then replace 'T' with '.'

        :-        # Then remove ':' and '-'

edited Jan 11 at 23:38

answered Jan 11 at 22:50

Jo King

21.5k248110

Perl 6, 57 50 bytes

{TR:d/T:-/./}o{substr ~DateTime.new($_/1e3): 2,15}

Try it online!

Takes the default stringification of a Datetime, in the format yyyy-mm-ddThh:mm:ssZ and modifies it to fit the output format. Perl 6 is in need of a date formatter method.

Explanation:

                       Datetime.new($_/1e3) # Create a date time

                      ~                 # Stringify it to the format yyyy-mm-ddThh:mm:ssZ

                                        # e.g. 2019-01-11T19:11:06.744000Z

               substr                      : 2,15  # Take the middle 15 characters

 {TR:d/T  /./}o   # Then replace 'T' with '.'

        :-        # Then remove ':' and '-'

edited Jan 11 at 23:38

answered Jan 11 at 22:50

Jo King

21.5k248110

edited Jan 11 at 23:38

answered Jan 11 at 22:50

Jo King

21.5k248110

answered Jan 11 at 22:50

Jo King

21.5k248110

answered Jan 11 at 22:50

Jo King

21.5k248110

add a comment |

C# (Visual C# Interactive Compiler), 67 61 60 bytes

n=>$"{new DateTime(1970,1,1).AddTicks(n*10000):yyMMdd.HHmm}"

For reasons unknown to me, DateTime.UnixEpoch doesn't work.

Try it online!

edited Jan 12 at 4:11

answered Jan 11 at 22:41

Embodiment of Ignorance

721116

1

$begingroup$
It seems UnixEpoch is only present in.Net Core 2.1+
$endgroup$
– digEmAll
Jan 12 at 12:19

add a comment |

C# (Visual C# Interactive Compiler), 67 61 60 bytes

n=>$"{new DateTime(1970,1,1).AddTicks(n*10000):yyMMdd.HHmm}"

For reasons unknown to me, DateTime.UnixEpoch doesn't work.

Try it online!

edited Jan 12 at 4:11

answered Jan 11 at 22:41

Embodiment of Ignorance

721116

1

$begingroup$
It seems UnixEpoch is only present in.Net Core 2.1+
$endgroup$
– digEmAll
Jan 12 at 12:19

add a comment |

C# (Visual C# Interactive Compiler), 67 61 60 bytes

n=>$"{new DateTime(1970,1,1).AddTicks(n*10000):yyMMdd.HHmm}"

For reasons unknown to me, DateTime.UnixEpoch doesn't work.

Try it online!

edited Jan 12 at 4:11

answered Jan 11 at 22:41

Embodiment of Ignorance

721116

C# (Visual C# Interactive Compiler), 67 61 60 bytes

n=>$"{new DateTime(1970,1,1).AddTicks(n*10000):yyMMdd.HHmm}"

For reasons unknown to me, DateTime.UnixEpoch doesn't work.

Try it online!

edited Jan 12 at 4:11

answered Jan 11 at 22:41

Embodiment of Ignorance

721116

edited Jan 12 at 4:11

answered Jan 11 at 22:41

Embodiment of Ignorance

721116

answered Jan 11 at 22:41

Embodiment of Ignorance

721116

answered Jan 11 at 22:41

Embodiment of Ignorance

721116

1

$begingroup$
It seems UnixEpoch is only present in.Net Core 2.1+
$endgroup$
– digEmAll
Jan 12 at 12:19

add a comment |

1

$begingroup$
It seems UnixEpoch is only present in.Net Core 2.1+
$endgroup$
– digEmAll
Jan 12 at 12:19

It seems UnixEpoch is only present in.Net Core 2.1+

– digEmAll
Jan 12 at 12:19

add a comment |

Javascript ES6, 76 66 bytes

x=>new Date(x).toJSON().slice(2,16).replace(/D/g,a=>a>'S'?'.':'')

Try it online

-10 bytes thanks to Shaggy!

x // timestamp

=>

new Date(x) // date object from timestamp

.toJSON() // same as .toISOString()

.slice(2,16) // cut off excess

.replace(/D/g, // match all non-digits

a // a is matched character

=>

a>'S'?'.' // if a is T (bigger than S is shorter) replace it with .

:''       // if it's not T, replace it with nothing

          // this way the dashes get removed and the dot gets put in the right place

) // end of replace

edited Jan 12 at 18:38

answered Jan 11 at 19:27

skiilaa

1638

3

$begingroup$
You may want to wait a day or so before answering your own questions next time.
$endgroup$
– fəˈnɛtɪk
Jan 11 at 20:11

$begingroup$
71 bytes
$endgroup$
– Luis felipe De jesus Munoz
Jan 11 at 20:16

$begingroup$
Alternative 71 bytes
$endgroup$
– Shaggy
Jan 11 at 20:33

1

$begingroup$
@LuisfelipeDejesusMunoz, that's different enough for you to post yourself.
$endgroup$
– Shaggy
Jan 11 at 21:49

$begingroup$
@LuisfelipeDejesusMunoz 66 bytes to tie you with Arnauld.
$endgroup$
– Shaggy
Jan 12 at 13:50

add a comment |

Javascript ES6, 76 66 bytes

x=>new Date(x).toJSON().slice(2,16).replace(/D/g,a=>a>'S'?'.':'')

Try it online

-10 bytes thanks to Shaggy!

x // timestamp

=>

new Date(x) // date object from timestamp

.toJSON() // same as .toISOString()

.slice(2,16) // cut off excess

.replace(/D/g, // match all non-digits

a // a is matched character

=>

a>'S'?'.' // if a is T (bigger than S is shorter) replace it with .

:''       // if it's not T, replace it with nothing

          // this way the dashes get removed and the dot gets put in the right place

) // end of replace

edited Jan 12 at 18:38

answered Jan 11 at 19:27

skiilaa

1638

3

$begingroup$
You may want to wait a day or so before answering your own questions next time.
$endgroup$
– fəˈnɛtɪk
Jan 11 at 20:11

$begingroup$
71 bytes
$endgroup$
– Luis felipe De jesus Munoz
Jan 11 at 20:16

$begingroup$
Alternative 71 bytes
$endgroup$
– Shaggy
Jan 11 at 20:33

1

$begingroup$
@LuisfelipeDejesusMunoz, that's different enough for you to post yourself.
$endgroup$
– Shaggy
Jan 11 at 21:49

$begingroup$
@LuisfelipeDejesusMunoz 66 bytes to tie you with Arnauld.
$endgroup$
– Shaggy
Jan 12 at 13:50

add a comment |

Javascript ES6, 76 66 bytes

x=>new Date(x).toJSON().slice(2,16).replace(/D/g,a=>a>'S'?'.':'')

Try it online

-10 bytes thanks to Shaggy!

x // timestamp

=>

new Date(x) // date object from timestamp

.toJSON() // same as .toISOString()

.slice(2,16) // cut off excess

.replace(/D/g, // match all non-digits

a // a is matched character

=>

a>'S'?'.' // if a is T (bigger than S is shorter) replace it with .

:''       // if it's not T, replace it with nothing

          // this way the dashes get removed and the dot gets put in the right place

) // end of replace

edited Jan 12 at 18:38

answered Jan 11 at 19:27

skiilaa

1638

Javascript ES6, 76 66 bytes

x=>new Date(x).toJSON().slice(2,16).replace(/D/g,a=>a>'S'?'.':'')

Try it online

-10 bytes thanks to Shaggy!

x // timestamp

=>

new Date(x) // date object from timestamp

.toJSON() // same as .toISOString()

.slice(2,16) // cut off excess

.replace(/D/g, // match all non-digits

a // a is matched character

=>

a>'S'?'.' // if a is T (bigger than S is shorter) replace it with .

:''       // if it's not T, replace it with nothing

          // this way the dashes get removed and the dot gets put in the right place

) // end of replace

edited Jan 12 at 18:38

answered Jan 11 at 19:27

skiilaa

1638

edited Jan 12 at 18:38

answered Jan 11 at 19:27

skiilaa

1638

answered Jan 11 at 19:27

skiilaa

1638

answered Jan 11 at 19:27

skiilaa

1638

3

$begingroup$
You may want to wait a day or so before answering your own questions next time.
$endgroup$
– fəˈnɛtɪk
Jan 11 at 20:11

$begingroup$
71 bytes
$endgroup$
– Luis felipe De jesus Munoz
Jan 11 at 20:16

$begingroup$
Alternative 71 bytes
$endgroup$
– Shaggy
Jan 11 at 20:33

1

$begingroup$
@LuisfelipeDejesusMunoz, that's different enough for you to post yourself.
$endgroup$
– Shaggy
Jan 11 at 21:49

$begingroup$
@LuisfelipeDejesusMunoz 66 bytes to tie you with Arnauld.
$endgroup$
– Shaggy
Jan 12 at 13:50

add a comment |

3

$begingroup$
You may want to wait a day or so before answering your own questions next time.
$endgroup$
– fəˈnɛtɪk
Jan 11 at 20:11

$begingroup$
71 bytes
$endgroup$
– Luis felipe De jesus Munoz
Jan 11 at 20:16

$begingroup$
Alternative 71 bytes
$endgroup$
– Shaggy
Jan 11 at 20:33

1

$begingroup$
@LuisfelipeDejesusMunoz, that's different enough for you to post yourself.
$endgroup$
– Shaggy
Jan 11 at 21:49

$begingroup$
@LuisfelipeDejesusMunoz 66 bytes to tie you with Arnauld.
$endgroup$
– Shaggy
Jan 12 at 13:50

You may want to wait a day or so before answering your own questions next time.

– fəˈnɛtɪk
Jan 11 at 20:11

71 bytes

– Luis felipe De jesus Munoz
Jan 11 at 20:16

Alternative 71 bytes

– Shaggy
Jan 11 at 20:33

@LuisfelipeDejesusMunoz, that's different enough for you to post yourself.

– Shaggy
Jan 11 at 21:49

@LuisfelipeDejesusMunoz 66 bytes to tie you with Arnauld.

– Shaggy
Jan 12 at 13:50

add a comment |

C (clang), 117 111 bytes

Thanks to @chux and @ceilingcat for the suggestions.

#import<time.h>

*l;f(long t){t/=1e3;printf("%02d%02d%02d.%02d%02d",5[l=gmtime(&t)]%100,l[4]+1,l[3],l[2],l[1]);}

Try it online!

edited Jan 14 at 12:15

answered Jan 12 at 3:26

ErikF

1,29917

$begingroup$
gmtime is short than localtime
$endgroup$
– chux
Jan 13 at 22:08

add a comment |

C (clang), 117 111 bytes

Thanks to @chux and @ceilingcat for the suggestions.

#import<time.h>

*l;f(long t){t/=1e3;printf("%02d%02d%02d.%02d%02d",5[l=gmtime(&t)]%100,l[4]+1,l[3],l[2],l[1]);}

Try it online!

edited Jan 14 at 12:15

answered Jan 12 at 3:26

ErikF

1,29917

$begingroup$
gmtime is short than localtime
$endgroup$
– chux
Jan 13 at 22:08

add a comment |

C (clang), 117 111 bytes

Thanks to @chux and @ceilingcat for the suggestions.

#import<time.h>

*l;f(long t){t/=1e3;printf("%02d%02d%02d.%02d%02d",5[l=gmtime(&t)]%100,l[4]+1,l[3],l[2],l[1]);}

Try it online!

edited Jan 14 at 12:15

answered Jan 12 at 3:26

ErikF

1,29917

C (clang), 117 111 bytes

Thanks to @chux and @ceilingcat for the suggestions.

#import<time.h>

*l;f(long t){t/=1e3;printf("%02d%02d%02d.%02d%02d",5[l=gmtime(&t)]%100,l[4]+1,l[3],l[2],l[1]);}

Try it online!

edited Jan 14 at 12:15

answered Jan 12 at 3:26

ErikF

1,29917

edited Jan 14 at 12:15

answered Jan 12 at 3:26

ErikF

1,29917

answered Jan 12 at 3:26

ErikF

1,29917

answered Jan 12 at 3:26

ErikF

1,29917

$begingroup$
gmtime is short than localtime
$endgroup$
– chux
Jan 13 at 22:08

add a comment |

$begingroup$
gmtime is short than localtime
$endgroup$
– chux
Jan 13 at 22:08

gmtime is short than localtime

– chux
Jan 13 at 22:08

add a comment |

Twig, 25 characters

{{d[:-3]|date('ymd.hi')}}

This is a template. Call it by including it and pass the Unix time as parameter d.

Sample usage:

{{include('datetime.twig', {'d': 1547233866744})}}

Try it on TwigFiddle

answered Jan 14 at 13:13

manatwork

16.3k43572

add a comment |

Twig, 25 characters

{{d[:-3]|date('ymd.hi')}}

This is a template. Call it by including it and pass the Unix time as parameter d.

Sample usage:

{{include('datetime.twig', {'d': 1547233866744})}}

Try it on TwigFiddle

answered Jan 14 at 13:13

manatwork

16.3k43572

add a comment |

Twig, 25 characters

{{d[:-3]|date('ymd.hi')}}

This is a template. Call it by including it and pass the Unix time as parameter d.

Sample usage:

{{include('datetime.twig', {'d': 1547233866744})}}

Try it on TwigFiddle

answered Jan 14 at 13:13

manatwork

16.3k43572

Twig, 25 characters

{{d[:-3]|date('ymd.hi')}}

This is a template. Call it by including it and pass the Unix time as parameter d.

Sample usage:

{{include('datetime.twig', {'d': 1547233866744})}}

Try it on TwigFiddle

answered Jan 14 at 13:13

manatwork

16.3k43572

answered Jan 14 at 13:13

manatwork

16.3k43572

answered Jan 14 at 13:13

manatwork

16.3k43572

answered Jan 14 at 13:13

manatwork

16.3k43572

add a comment |

JavaScript, 64 bytes

n=>'2356891911121415'.replace(/1?./g,x=>new Date(n).toJSON()[x])

Try it online!

answered Jan 15 at 2:36

tsh

8,61511547

$begingroup$
aw, so close
$endgroup$
– ASCII-only
Jan 15 at 5:47

$begingroup$
also 64
$endgroup$
– ASCII-only
Jan 15 at 5:48

add a comment |

JavaScript, 64 bytes

n=>'2356891911121415'.replace(/1?./g,x=>new Date(n).toJSON()[x])

Try it online!

answered Jan 15 at 2:36

tsh

8,61511547

$begingroup$
aw, so close
$endgroup$
– ASCII-only
Jan 15 at 5:47

$begingroup$
also 64
$endgroup$
– ASCII-only
Jan 15 at 5:48

add a comment |

JavaScript, 64 bytes

n=>'2356891911121415'.replace(/1?./g,x=>new Date(n).toJSON()[x])

Try it online!

answered Jan 15 at 2:36

tsh

8,61511547

JavaScript, 64 bytes

n=>'2356891911121415'.replace(/1?./g,x=>new Date(n).toJSON()[x])

Try it online!

answered Jan 15 at 2:36

tsh

8,61511547

answered Jan 15 at 2:36

tsh

8,61511547

answered Jan 15 at 2:36

tsh

8,61511547

answered Jan 15 at 2:36

tsh

8,61511547

$begingroup$
aw, so close
$endgroup$
– ASCII-only
Jan 15 at 5:47

$begingroup$
also 64
$endgroup$
– ASCII-only
Jan 15 at 5:48

add a comment |

$begingroup$
aw, so close
$endgroup$
– ASCII-only
Jan 15 at 5:47

$begingroup$
also 64
$endgroup$
– ASCII-only
Jan 15 at 5:48

aw, so close

– ASCII-only
Jan 15 at 5:47

also 64

– ASCII-only
Jan 15 at 5:48

add a comment |

jq, 33 characters

(30 characters code + 3 characters command line option)

./1000|strftime("%y%m%d.%H%M")

Sample run:

bash-4.4$ jq -r './1000|strftime("%y%m%d.%H%M")' <<< 1547233866744

190111.1911

Try it online!

answered Jan 11 at 20:20

manatwork

16.3k43572

3

$begingroup$
You don't need to count command-line flags anymore.
$endgroup$
– AdmBorkBork
Jan 11 at 20:28

$begingroup$
Oops. Good to know. Thank you @AdmBorkBork.
$endgroup$
– manatwork
Jan 11 at 20:36

$begingroup$
1000 -> 1e3
$endgroup$
– Shaggy
Jan 14 at 20:16

add a comment |

jq, 33 characters

(30 characters code + 3 characters command line option)

./1000|strftime("%y%m%d.%H%M")

Sample run:

bash-4.4$ jq -r './1000|strftime("%y%m%d.%H%M")' <<< 1547233866744

190111.1911

Try it online!

answered Jan 11 at 20:20

manatwork

16.3k43572

3

$begingroup$
You don't need to count command-line flags anymore.
$endgroup$
– AdmBorkBork
Jan 11 at 20:28

$begingroup$
Oops. Good to know. Thank you @AdmBorkBork.
$endgroup$
– manatwork
Jan 11 at 20:36

$begingroup$
1000 -> 1e3
$endgroup$
– Shaggy
Jan 14 at 20:16

add a comment |

jq, 33 characters

(30 characters code + 3 characters command line option)

./1000|strftime("%y%m%d.%H%M")

Sample run:

bash-4.4$ jq -r './1000|strftime("%y%m%d.%H%M")' <<< 1547233866744

190111.1911

Try it online!

answered Jan 11 at 20:20

manatwork

16.3k43572

jq, 33 characters

(30 characters code + 3 characters command line option)

./1000|strftime("%y%m%d.%H%M")

Sample run:

bash-4.4$ jq -r './1000|strftime("%y%m%d.%H%M")' <<< 1547233866744

190111.1911

Try it online!

answered Jan 11 at 20:20

manatwork

16.3k43572

answered Jan 11 at 20:20

manatwork

16.3k43572

answered Jan 11 at 20:20

manatwork

16.3k43572

answered Jan 11 at 20:20

manatwork

16.3k43572

3

$begingroup$
You don't need to count command-line flags anymore.
$endgroup$
– AdmBorkBork
Jan 11 at 20:28

$begingroup$
Oops. Good to know. Thank you @AdmBorkBork.
$endgroup$
– manatwork
Jan 11 at 20:36

$begingroup$
1000 -> 1e3
$endgroup$
– Shaggy
Jan 14 at 20:16

add a comment |

3

$begingroup$
You don't need to count command-line flags anymore.
$endgroup$
– AdmBorkBork
Jan 11 at 20:28

$begingroup$
Oops. Good to know. Thank you @AdmBorkBork.
$endgroup$
– manatwork
Jan 11 at 20:36

$begingroup$
1000 -> 1e3
$endgroup$
– Shaggy
Jan 14 at 20:16

You don't need to count command-line flags anymore.

– AdmBorkBork
Jan 11 at 20:28

Oops. Good to know. Thank you @AdmBorkBork.

– manatwork
Jan 11 at 20:36

1000 -> 1e3

– Shaggy
Jan 14 at 20:16

add a comment |

ksh, 36 bytes

printf "%(%y%m%d.%H%M)T" $(($1/1e3))

Try it online!

Thanks to Jo King for 15 bytes saved

edited Jan 12 at 2:55

answered Jan 12 at 1:56

Sergiy Kolodyazhnyy

331110

$begingroup$
The same in Bash would be just 35 characters: Try it online!
$endgroup$
– manatwork
Jan 13 at 15:25

add a comment |

ksh, 36 bytes

printf "%(%y%m%d.%H%M)T" $(($1/1e3))

Try it online!

Thanks to Jo King for 15 bytes saved

edited Jan 12 at 2:55

answered Jan 12 at 1:56

Sergiy Kolodyazhnyy

331110

$begingroup$
The same in Bash would be just 35 characters: Try it online!
$endgroup$
– manatwork
Jan 13 at 15:25

add a comment |

ksh, 36 bytes

printf "%(%y%m%d.%H%M)T" $(($1/1e3))

Try it online!

Thanks to Jo King for 15 bytes saved

edited Jan 12 at 2:55

answered Jan 12 at 1:56

Sergiy Kolodyazhnyy

331110

ksh, 36 bytes

printf "%(%y%m%d.%H%M)T" $(($1/1e3))

Try it online!

Thanks to Jo King for 15 bytes saved

edited Jan 12 at 2:55

answered Jan 12 at 1:56

Sergiy Kolodyazhnyy

331110

edited Jan 12 at 2:55

answered Jan 12 at 1:56

Sergiy Kolodyazhnyy

331110

answered Jan 12 at 1:56

Sergiy Kolodyazhnyy

331110

answered Jan 12 at 1:56

Sergiy Kolodyazhnyy

331110

$begingroup$
The same in Bash would be just 35 characters: Try it online!
$endgroup$
– manatwork
Jan 13 at 15:25

add a comment |

$begingroup$
The same in Bash would be just 35 characters: Try it online!
$endgroup$
– manatwork
Jan 13 at 15:25

The same in Bash would be just 35 characters: Try it online!

– manatwork
Jan 13 at 15:25

add a comment |

MediaWiki, 46 bytes

{{#time:ymd.Hi|@{{#expr:floor({{{1}}}/1e3)}}}}

answered Jan 13 at 11:19

tsh

8,61511547

add a comment |

MediaWiki, 46 bytes

{{#time:ymd.Hi|@{{#expr:floor({{{1}}}/1e3)}}}}

answered Jan 13 at 11:19

tsh

8,61511547

add a comment |

MediaWiki, 46 bytes

{{#time:ymd.Hi|@{{#expr:floor({{{1}}}/1e3)}}}}

answered Jan 13 at 11:19

tsh

8,61511547

MediaWiki, 46 bytes

{{#time:ymd.Hi|@{{#expr:floor({{{1}}}/1e3)}}}}

answered Jan 13 at 11:19

tsh

8,61511547

answered Jan 13 at 11:19

tsh

8,61511547

answered Jan 13 at 11:19

tsh

8,61511547

answered Jan 13 at 11:19

tsh

8,61511547

add a comment |

Java 8, 78 bytes

n->new java.text.SimpleDateFormat("yyMMdd.HHmm").format(new java.util.Date(n))

Try it online.

Explanation:

n->                       // Method with long parameter and String return-type

  new java.text.SimpleDateFormat("yyMMdd.HHmm")

                          //  Create the formatter

   .format(               //  Format the date to a String in this format and return it:

     new java.util.Date(  //   Create a new Date

      n))                 //   With the input-long as timestamp

answered Jan 15 at 8:24

Kevin Cruijssen

36.8k555193

add a comment |

Java 8, 78 bytes

n->new java.text.SimpleDateFormat("yyMMdd.HHmm").format(new java.util.Date(n))

Try it online.

Explanation:

n->                       // Method with long parameter and String return-type

  new java.text.SimpleDateFormat("yyMMdd.HHmm")

                          //  Create the formatter

   .format(               //  Format the date to a String in this format and return it:

     new java.util.Date(  //   Create a new Date

      n))                 //   With the input-long as timestamp

answered Jan 15 at 8:24

Kevin Cruijssen

36.8k555193

add a comment |

Java 8, 78 bytes

n->new java.text.SimpleDateFormat("yyMMdd.HHmm").format(new java.util.Date(n))

Try it online.

Explanation:

n->                       // Method with long parameter and String return-type

  new java.text.SimpleDateFormat("yyMMdd.HHmm")

                          //  Create the formatter

   .format(               //  Format the date to a String in this format and return it:

     new java.util.Date(  //   Create a new Date

      n))                 //   With the input-long as timestamp

answered Jan 15 at 8:24

Kevin Cruijssen

36.8k555193

Java 8, 78 bytes

n->new java.text.SimpleDateFormat("yyMMdd.HHmm").format(new java.util.Date(n))

Try it online.

Explanation:

n->                       // Method with long parameter and String return-type

  new java.text.SimpleDateFormat("yyMMdd.HHmm")

                          //  Create the formatter

   .format(               //  Format the date to a String in this format and return it:

     new java.util.Date(  //   Create a new Date

      n))                 //   With the input-long as timestamp

answered Jan 15 at 8:24

Kevin Cruijssen

36.8k555193

answered Jan 15 at 8:24

Kevin Cruijssen

36.8k555193

answered Jan 15 at 8:24

Kevin Cruijssen

36.8k555193

answered Jan 15 at 8:24

Kevin Cruijssen

36.8k555193

add a comment |

If this is an answer to a challenge…

…Be sure to follow the challenge specification. However, please refrain from exploiting obvious loopholes. Answers abusing any of the standard loopholes are considered invalid. If you think a specification is unclear or underspecified, comment on the question instead.

…Try to optimize your score. For instance, answers to code-golf challenges should attempt to be as short as possible. You can always include a readable version of the code in addition to the competitive one.
Explanations of your answer make it more interesting to read and are very much encouraged.

…Include a short header which indicates the language(s) of your code and its score, as defined by the challenge.

More generally…

…Please make sure to answer the question and provide sufficient detail.

…Avoid asking for help, clarification or responding to other answers (use comments instead).

draft saved

draft discarded

If this is an answer to a challenge…

…Be sure to follow the challenge specification. However, please refrain from exploiting obvious loopholes. Answers abusing any of the standard loopholes are considered invalid. If you think a specification is unclear or underspecified, comment on the question instead.

…Try to optimize your score. For instance, answers to code-golf challenges should attempt to be as short as possible. You can always include a readable version of the code in addition to the competitive one.
Explanations of your answer make it more interesting to read and are very much encouraged.

…Include a short header which indicates the language(s) of your code and its score, as defined by the challenge.

More generally…

…Please make sure to answer the question and provide sufficient detail.

…Avoid asking for help, clarification or responding to other answers (use comments instead).

draft saved

draft discarded

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This page is only for reference, If you need detailed information, please check here

Unix timestamp to datetime string

Rules

Example

Rules

Example

Rules

Example

Rules

Example

21 Answers 21

Japt v1.4.5, 19 16 bytes

Explanation

Notes / Tips

JavaScript (ES6), 65 bytes

How?

JavaScript (ES6), 66 bytes

How?

Bash + coreutils, 29 bytes

PHP, 40 32 31 bytes

MATL, 28 bytes

Explanation

GNU AWK, 34 33 characters

PowerShell, 59 58 bytes

Perl 6, 111 89 87 bytes

Explanation:

Python 2, 64 bytes

R, 58 56 bytes

C (gcc) (32-bit, little endian), 67 bytes

Perl 6, 57 50 bytes

Explanation:

C# (Visual C# Interactive Compiler), 67 61 60 bytes

Javascript ES6, 76 66 bytes

C (clang), 117 111 bytes

Twig, 25 characters

JavaScript, 64 bytes

jq, 33 characters

ksh, 36 bytes

MediaWiki, 46 bytes

Java 8, 78 bytes

Your Answer

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21 Answers 21

21 Answers 21

Japt v1.4.5, 19 16 bytes

Explanation

Notes / Tips

Japt v1.4.5, 19 16 bytes

Explanation

Notes / Tips

Japt v1.4.5, 19 16 bytes

Explanation

Notes / Tips

Japt v1.4.5, 19 16 bytes

Explanation

Notes / Tips

JavaScript (ES6), 65 bytes

How?

JavaScript (ES6), 66 bytes

How?

JavaScript (ES6), 65 bytes

How?

JavaScript (ES6), 66 bytes

How?

JavaScript (ES6), 65 bytes

How?

JavaScript (ES6), 66 bytes

How?

JavaScript (ES6), 65 bytes

How?

JavaScript (ES6), 66 bytes

How?

Bash + coreutils, 29 bytes

Bash + coreutils, 29 bytes

Bash + coreutils, 29 bytes

Bash + coreutils, 29 bytes

PHP, 40 32 31 bytes

PHP, 40 32 31 bytes

PHP, 40 32 31 bytes

21 Answers
21

21 Answers
21

21 Answers
21