Unix & Linux Stack Exchange is a question and answer site for users of Linux, FreeBSD and other Un*x-like operating systems. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Say I have a file:

# file: 'test.txt'
foobar bash 1
foobar happy

I only want to know what words appear after "foobar", so I can use this regex:

"foobar \(\w\+\)"

The parenthesis indicate that I have a special interest in the word right after foobar. But when I do a grep "foobar \(\w\+\)" test.txt, I get the entire lines that match the entire regex, rather than just "the word after foobar":

foobar bash 1
foobar happy

I would much prefer that the output of that command looked like this:


Is there a way to tell grep to only output the items that match the grouping (or a specific grouping) in a regular expression?

share|improve this question
up vote 121 down vote accepted

GNU grep has the -P option for perl-style regexes, and the -o option to print only what matches the pattern. These can be combined using look-around assertions (described under Extended Patterns in the perlre manpage) to remove part of the grep pattern from what is determined to have matched for the purposes of -o.

$ grep -oP 'foobar \K\w+' test.txt

The \K is the short-form (and more efficient form) of (?<=pattern) which you use as a zero-width look-behind assertion before the text you want to output. (?=pattern) can be used as a zero-width look-ahead assertion after the text you want to output.

For instance, if you wanted to match the word between foo and bar, you could use:

$ grep -oP 'foo \K\w+(?= bar)' test.txt

or (for symmetry)

$ grep -oP '(?<=foo )\w+(?= bar)' test.txt
share|improve this answer
How you do it if your regex has more than a grouping? (as the title implied?) – barracel Mar 21 '13 at 7:52
@barracel: I don't believe you can. Time for sed(1) – camh Mar 22 '13 at 22:51
@camh I have just tested that grep -oP 'foobar \K\w+' test.txt outputs nothing with the OP's test.txt. The grep version is 2.5.1. What could be wrong ? O_O – Xichen Li Jul 24 '14 at 14:19
@XichenLi: I can't say. I just built v2.5.1 of grep (it's pretty old - from 2006) and it worked for me. – camh Jul 25 '14 at 10:18

Standard grep can't do this, but recent versions of GNU grep can. You can turn to sed, awk or perl. Here are a few examples that do what you want on your sample input; they behave slightly differently in corner cases.

Replace foobar word other stuff by word, print only if a replacement is done.

sed -n -e 's/^foobar \([[:alnum:]]\+\).*/\1/p'

If the first word is foobar, print the second word.

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

Strip foobar if it's the first word, and skip the line otherwise; then strip everything after the first whitespace and print.

perl -lne 's/^foobar\s+// or next; s/\s.*//; print'
share|improve this answer
Awesome! I thought I may be able to do this with sed, but I haven't used it before and was hoping I could use my familiar grep. But the syntax for these commands actually looks very familiar now that I am familiar with vim-style search & replace + regexes. Thanks a ton. – Cory Klein May 19 '11 at 23:51
Not true, Gilles. See my answer for a GNU grep solution. – camh May 20 '11 at 1:33
@camh: Ah, I didn't know GNU grep now had full PCRE support. I've corrected my answer, thanks. – Gilles May 20 '11 at 7:14
This answer is especially useful for embedded Linux since Busybox grep doesn't have PCRE support. – Craig McQueen Mar 17 at 0:12

Well, if you know that foobar is always the first word or the line, then you can use cut. Like so:

grep "foobar" test.file | cut -d" " -f2
share|improve this answer
The -o switch on grep is widely implemented (moreso than the Gnu grep extensions), so doing grep -o "foobar" test.file | cut -d" " -f2 will increase the effectiveness of this solution, which is more portable than using lookbehind assertions. – dubiousjim Apr 19 '12 at 21:04

If PCRE is not supported you can achieve the same result with two invocations of grep. For example to grab the word after foobar do this:

<test.txt grep -o 'foobar  *[^ ]*' | grep -o '[^ ]*$'

This can be expanded to an arbitrary word after foobar like this (with EREs for readability):

<test.txt egrep -o 'foobar +([^ ]+ +){'$i'}[^ ]+' | grep -o '[^ ]*$'



Note the index i is zero-based.

share|improve this answer
    sed -n "s/^.*foobar\s*\(\S*\).*$/\1/p"

-n     suppress printing
s      substitute
^.*    anything before foobar
foobar initial search match
\s*    any white space character (space)
\(     start capture group
\S*    capture any non-white space character (word)
\)     end capture group
.*$    anything after the capture group
\1     substitute everything with the 1st capture group
p      print it
share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.