Can GNU Grep output a selected group?

Question

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?

Answer 1

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"

Answer 2

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 "(?<=')[^']+(?=')"

Answer 3

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.