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Is it possible to use GNU grep to get a matched group from an expression?
Example:
echo "foo 'bar'" | grep -oE "'([^']+)'"
Which would output "'bar'". But I would like to get just "bar", without having to send it through grep one more time (ie. get the matched group). Is that possible?
You can use sed for this. On BSD sed:
echo "foo 'bar'" | sed -E "s/.*'([^']+)'.*/\\1/"
Or, without the -E option:
sed "s/.*'\([^']\+\)'.*/\1/"
This doesn't work for multiline input. For that you need:
sed -n "s/.*'\([^']\+\)'.*/\1/p"
grep's -E: "Interpret regular expressions as extended (modern) regular expressions rather than basic regular expressions (BRE's). The re_format(7) manual page fully describes both formats."
( ) + use \( \) \+: This is effectively the same: sed "s/.*'\([^']\+\)'.*/\1/"
sed -n "s/.*'\([^']\+\)'.*/\1/p"
Oct 10, 2012 at 18:15
While grep can't output a specific group, you can use lookahead and behind assertions to achieve what your after:
echo "foo 'bar'" | grep -Po "(?<=')[^']+(?=')"
grep -P is not available on all platforms. But if it is, using lookahead/behind is a very nice way of solving the problem.
Jun 13, 2012 at 13:16
You can use \K to reset and discard the left hand match text along with a lookahead which is not added to the match text:
$ echo "foo 'bar'" | grep -oP "'\K[^']+(?=')"
bar
GNU grep only.
