Awk Print Line After A Matching /regex/

by on May 13, 2009 · 11 comments· LAST UPDATED May 13, 2009

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How do I a print line after matching /regex/ using awk under Linux / UNIX operating systems? I'd like print the line immediately after a line that matches /regex/ but not the line that matches itself.

You can use awk, sed and any other tool for the same. Here is awk syntax:
awk '/regex/ { getline; print $0 }' /path/to/file
awk -F: '/regex/ { getline; print $0 }' /path/to/file

getline used to set $0 from next input record; it also set NF, NR, FNR. For example, match 'Registrar:' string and print actual registrar name from whois look up:
whois bbc.co.uk | awk -F: '/Registrar:/ && $0 != "" { getline; print $0}'
Sample output:

        British Broadcasting Corporation [Tag = BBC]
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{ 11 comments… read them below or add one }

1 Hai Vu May 13, 2009 at 4:03 pm

I suggest to use print instead of print $0: it is shorter.

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2 Nikhil May 14, 2009 at 9:06 am

$0 != “” looks like a redundant check since it is still the line matching the regex and not the next line.

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3 GByte May 14, 2009 at 9:15 am

How about:
$pgrep -A 1 regexp | tail -n 1
?

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4 Julio A. Cartaya May 16, 2009 at 10:17 am

Simply saying:
$ awk ‘/regexp/’ /pathto/file
will do the trick: the default action in awk is printing.

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5 nyxx May 16, 2009 at 6:26 pm

Reducing awk code for a while we’ll get tha same
$ whois bbc.co.uk | awk -F: '/Registrar:/ { getline; print }'
British Broadcasting Corporation [Tag = BBC]

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6 Ajay June 8, 2009 at 3:12 am

You can us grep as well:
whois bbc.co.uk | grep -A 1 Registrar| grep -v Registrar

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7 paddu July 1, 2011 at 11:30 am

I have a line abc=xyz
i want to find a pattern “abc=” and if exist print xyz
can some one please help

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8 royi January 7, 2012 at 12:15 am

awk -F “=” ‘{if ($1 like “abc”) print $2 }’

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9 Philippe Petrinko April 25, 2013 at 8:50 am

Hi, nice topic.

By the way,

1) using [-F] parameter is totally useless here, there is no point in defining [Field separator] as “:”.

2) As Nikhil stated, test on [$0] is also useless.

3) You may want to get rid of heading 8 space characters, using [substr]

So this code will do the trick:

whois bbc.co.uk | awk ‘/Registrar:/ { getline; print substr($0,9)}’

Thanks to change your article accordingly,

– Philippe

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10 Martin Heimsoth March 13, 2014 at 7:30 am

Hello everyone,
I was looking for such a solution, and I think I found something better than Philippe proposes. The thing is that very often one does not know how long the remaining string to be printed will be. So here is what I did:

awk ‘/REGEX/ {gsub(“REGEX”,””);print;}’ /path/to/file

Note that the regular expression in the first argument of gsub has to be the same as the search pattern, except for some special characters which might need an escape sequence.

Sorry for showing up so late :)

- Martin

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11 Philippe Petrinko March 13, 2014 at 8:51 am

@Martin Heimsoth

I would be glad to learn something new and better than solution I know, but I don’t get your point.

I think you may have not understand what I coded.
My solution does not depend on length of remaining string.

” substr($0,9) ” cuts a line from 8 heading characters, and this will work as designed whatever length of remaining string may be.

There is no problem regarding length of printed line, CMIIAW

– Philippe

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