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 so far... add one ↓
🐧 11 comments so far... add one ↓
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I suggest to use print instead of print $0: it is shorter.
$0 != “” looks like a redundant check since it is still the line matching the regex and not the next line.
How about:
$pgrep -A 1 regexp | tail -n 1
?
Simply saying:
$ awk ‘/regexp/’ /pathto/file
will do the trick: the default action in awk is printing.
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]
You can us grep as well:
whois bbc.co.uk | grep -A 1 Registrar| grep -v Registrar
I have a line abc=xyz
i want to find a pattern “abc=” and if exist print xyz
can some one please help
awk -F “=” ‘{if ($1 like “abc”) print $2 }’
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
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
@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