grep specific number from lines 5 but not 25 or 52 and so on

I need to find a user for a userid. The return from the authentication system is as follows. Now with a bash script I need to extract the line with the exact number. Not 5 and 25 ..

------------

ID  LOGIN   

------------

28  user1

25  user2 

5   user3

If I use grep 5 I do get 2 lines but I need the line with "5".

Any ideas ?

edited Jan 9 at 12:15

don_crissti

50.3k15134162

asked Jan 9 at 12:13

Dave.

241

add a comment |

I need to find a user for a userid. The return from the authentication system is as follows. Now with a bash script I need to extract the line with the exact number. Not 5 and 25 ..

------------

ID  LOGIN   

------------

28  user1

25  user2 

5   user3

If I use grep 5 I do get 2 lines but I need the line with "5".

Any ideas ?

edited Jan 9 at 12:15

don_crissti

50.3k15134162

asked Jan 9 at 12:13

Dave.

241

add a comment |

I need to find a user for a userid. The return from the authentication system is as follows. Now with a bash script I need to extract the line with the exact number. Not 5 and 25 ..

------------

ID  LOGIN   

------------

28  user1

25  user2 

5   user3

If I use grep 5 I do get 2 lines but I need the line with "5".

Any ideas ?

edited Jan 9 at 12:15

don_crissti

50.3k15134162

asked Jan 9 at 12:13

Dave.

241

I need to find a user for a userid. The return from the authentication system is as follows. Now with a bash script I need to extract the line with the exact number. Not 5 and 25 ..

------------

ID  LOGIN   

------------

28  user1

25  user2 

5   user3

If I use grep 5 I do get 2 lines but I need the line with "5".

Any ideas ?

text-processing

edited Jan 9 at 12:15

don_crissti

50.3k15134162

asked Jan 9 at 12:13

Dave.

241

edited Jan 9 at 12:15

don_crissti

50.3k15134162

asked Jan 9 at 12:13

Dave.

241

edited Jan 9 at 12:15

don_crissti

50.3k15134162

edited Jan 9 at 12:15

don_crissti

50.3k15134162

edited Jan 9 at 12:15

don_crissti

50.3k15134162

asked Jan 9 at 12:13

Dave.

241

asked Jan 9 at 12:13

Dave.

241

asked Jan 9 at 12:13

Dave.

241

add a comment |

3 Answers
3

active

oldest

votes

There are several ways of doing it.

IMHO the best way is to use awk, which is useful when dealing with fields.

If you want a grep based solution, I would do:

grep -w '^5'

The -w tells grep to match the exact word, so this will not match "52". The "^" tells grep to search the 5 at the beginning of the line, which will fail if there are e.g. leading spaces.

The awk solution would look like:

awk '$1 == 5'

If you want only the username, which is the second column:

awk '$1 == 5 {print $2}'

If you're searching for a string and not a numeric value, enclose the string in double quotes:

awk '$1 == "abc" {print $2}'

edited Jan 12 at 19:22

Jeff Schaller

39.5k1054126

answered Jan 9 at 12:24

wurtel

10k11425

1

thanks I have choosen the awk way. awk '$1 == 5 {print $2}' that worked.

– Dave.
Jan 9 at 14:05

2

@Dave. Since this answer worked for you, please consider clicking the checkbox beside it to signal to future readers it answered your question.

– bishop
Jan 9 at 20:59

add a comment |

You could try with a regex (first char in line) and including the space:

grep -E "^5 "

answered Jan 9 at 12:21

rbrtflr

613

4

This would be better as grep -E "^5s" in case there's a tab and not a space.

– terdon♦
Jan 9 at 13:50

add a comment |

Try the following:

query-auth-system | grep "^5\>"

^: means "match at start of line

\>: matches a word-boundary. So it will match 5, but not 50.

answered Jan 9 at 12:19

Ralf

3277

2

That will also match lines starting with 5:, 5%, 5| etc.

– terdon♦
Jan 9 at 13:49

add a comment |

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

There are several ways of doing it.

IMHO the best way is to use awk, which is useful when dealing with fields.

If you want a grep based solution, I would do:

grep -w '^5'

The -w tells grep to match the exact word, so this will not match "52". The "^" tells grep to search the 5 at the beginning of the line, which will fail if there are e.g. leading spaces.

The awk solution would look like:

awk '$1 == 5'

If you want only the username, which is the second column:

awk '$1 == 5 {print $2}'

If you're searching for a string and not a numeric value, enclose the string in double quotes:

awk '$1 == "abc" {print $2}'

edited Jan 12 at 19:22

Jeff Schaller

39.5k1054126

answered Jan 9 at 12:24

wurtel

10k11425

1

thanks I have choosen the awk way. awk '$1 == 5 {print $2}' that worked.

– Dave.
Jan 9 at 14:05

2

@Dave. Since this answer worked for you, please consider clicking the checkbox beside it to signal to future readers it answered your question.

– bishop
Jan 9 at 20:59

add a comment |

There are several ways of doing it.

IMHO the best way is to use awk, which is useful when dealing with fields.

If you want a grep based solution, I would do:

grep -w '^5'

The -w tells grep to match the exact word, so this will not match "52". The "^" tells grep to search the 5 at the beginning of the line, which will fail if there are e.g. leading spaces.

The awk solution would look like:

awk '$1 == 5'

If you want only the username, which is the second column:

awk '$1 == 5 {print $2}'

If you're searching for a string and not a numeric value, enclose the string in double quotes:

awk '$1 == "abc" {print $2}'

edited Jan 12 at 19:22

Jeff Schaller

39.5k1054126

answered Jan 9 at 12:24

wurtel

10k11425

1

thanks I have choosen the awk way. awk '$1 == 5 {print $2}' that worked.

– Dave.
Jan 9 at 14:05

2

@Dave. Since this answer worked for you, please consider clicking the checkbox beside it to signal to future readers it answered your question.

– bishop
Jan 9 at 20:59

add a comment |

There are several ways of doing it.

IMHO the best way is to use awk, which is useful when dealing with fields.

If you want a grep based solution, I would do:

grep -w '^5'

The -w tells grep to match the exact word, so this will not match "52". The "^" tells grep to search the 5 at the beginning of the line, which will fail if there are e.g. leading spaces.

The awk solution would look like:

awk '$1 == 5'

If you want only the username, which is the second column:

awk '$1 == 5 {print $2}'

If you're searching for a string and not a numeric value, enclose the string in double quotes:

awk '$1 == "abc" {print $2}'

edited Jan 12 at 19:22

Jeff Schaller

39.5k1054126

answered Jan 9 at 12:24

wurtel

10k11425

There are several ways of doing it.

IMHO the best way is to use awk, which is useful when dealing with fields.

If you want a grep based solution, I would do:

grep -w '^5'

The -w tells grep to match the exact word, so this will not match "52". The "^" tells grep to search the 5 at the beginning of the line, which will fail if there are e.g. leading spaces.

The awk solution would look like:

awk '$1 == 5'

If you want only the username, which is the second column:

awk '$1 == 5 {print $2}'

If you're searching for a string and not a numeric value, enclose the string in double quotes:

awk '$1 == "abc" {print $2}'

edited Jan 12 at 19:22

Jeff Schaller

39.5k1054126

answered Jan 9 at 12:24

wurtel

10k11425

edited Jan 12 at 19:22

Jeff Schaller

39.5k1054126

edited Jan 12 at 19:22

Jeff Schaller

39.5k1054126

edited Jan 12 at 19:22

Jeff Schaller

39.5k1054126

answered Jan 9 at 12:24

wurtel

10k11425

answered Jan 9 at 12:24

wurtel

10k11425

answered Jan 9 at 12:24

wurtel

10k11425

1

thanks I have choosen the awk way. awk '$1 == 5 {print $2}' that worked.

– Dave.
Jan 9 at 14:05

2

@Dave. Since this answer worked for you, please consider clicking the checkbox beside it to signal to future readers it answered your question.

– bishop
Jan 9 at 20:59

add a comment |

1

thanks I have choosen the awk way. awk '$1 == 5 {print $2}' that worked.

– Dave.
Jan 9 at 14:05

2

@Dave. Since this answer worked for you, please consider clicking the checkbox beside it to signal to future readers it answered your question.

– bishop
Jan 9 at 20:59

thanks I have choosen the awk way. awk '$1 == 5 {print $2}' that worked.

– Dave.
Jan 9 at 14:05

@Dave. Since this answer worked for you, please consider clicking the checkbox beside it to signal to future readers it answered your question.

– bishop
Jan 9 at 20:59

add a comment |

You could try with a regex (first char in line) and including the space:

grep -E "^5 "

answered Jan 9 at 12:21

rbrtflr

613

4

This would be better as grep -E "^5s" in case there's a tab and not a space.

– terdon♦
Jan 9 at 13:50

add a comment |

You could try with a regex (first char in line) and including the space:

grep -E "^5 "

answered Jan 9 at 12:21

rbrtflr

613

4

This would be better as grep -E "^5s" in case there's a tab and not a space.

– terdon♦
Jan 9 at 13:50

add a comment |

You could try with a regex (first char in line) and including the space:

grep -E "^5 "

answered Jan 9 at 12:21

rbrtflr

613

You could try with a regex (first char in line) and including the space:

grep -E "^5 "

answered Jan 9 at 12:21

rbrtflr

613

answered Jan 9 at 12:21

rbrtflr

613

answered Jan 9 at 12:21

rbrtflr

613

answered Jan 9 at 12:21

rbrtflr

613

4

This would be better as grep -E "^5s" in case there's a tab and not a space.

– terdon♦
Jan 9 at 13:50

add a comment |

4

This would be better as grep -E "^5s" in case there's a tab and not a space.

– terdon♦
Jan 9 at 13:50

This would be better as grep -E "^5s" in case there's a tab and not a space.

– terdon♦
Jan 9 at 13:50

add a comment |

Try the following:

query-auth-system | grep "^5\>"

^: means "match at start of line

\>: matches a word-boundary. So it will match 5, but not 50.

answered Jan 9 at 12:19

Ralf

3277

2

That will also match lines starting with 5:, 5%, 5| etc.

– terdon♦
Jan 9 at 13:49

add a comment |

Try the following:

query-auth-system | grep "^5\>"

^: means "match at start of line

\>: matches a word-boundary. So it will match 5, but not 50.

answered Jan 9 at 12:19

Ralf

3277

2

That will also match lines starting with 5:, 5%, 5| etc.

– terdon♦
Jan 9 at 13:49

add a comment |

Try the following:

query-auth-system | grep "^5\>"

^: means "match at start of line

\>: matches a word-boundary. So it will match 5, but not 50.

answered Jan 9 at 12:19

Ralf

3277

Try the following:

query-auth-system | grep "^5\>"

^: means "match at start of line

\>: matches a word-boundary. So it will match 5, but not 50.

answered Jan 9 at 12:19

Ralf

3277

answered Jan 9 at 12:19

Ralf

3277

answered Jan 9 at 12:19

Ralf

3277

answered Jan 9 at 12:19

Ralf

3277

2

That will also match lines starting with 5:, 5%, 5| etc.

– terdon♦
Jan 9 at 13:49

add a comment |

2

That will also match lines starting with 5:, 5%, 5| etc.

– terdon♦
Jan 9 at 13:49

That will also match lines starting with 5:, 5%, 5| etc.

– terdon♦
Jan 9 at 13:49

add a comment |

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