Show only part of the line found by grep (but not the part that matches the pattern)

The canonical tool for that would be sed.

sed -n -e 's/^.*stalled: //p'

Detailed explanation:

• -n means not to print anything by default.
• -e is followed by a sed command.
• s is the pattern replacement command.
• The regular expression ^.*stalled: matches the pattern you're looking for, plus any preceding text (.* meaning any text, with an initial ^ to say that the match begins at the beginning of the line). Note that if stalled: occurs several times on the line, this will match the last occurrence.
• The match, i.e. everything on the line up to stalled:, is replaced by the empty string (i.e. deleted).
• The final p means to print the transformed line.

If you want to retain the matching portion, use a backreference: \1 in the replacement part designates what is inside a group \(…\) in the pattern. Here, you could write stalled: again in the replacement part; this feature is useful when the pattern you're looking for is more general than a simple string.

sed -n -e 's/^.*\(stalled: \)/\1/p'

Sometimes you'll want to remove the portion of the line after the match. You can include it in the match by including .*$ at the end of the pattern (any text .* followed by the end of the line $). Unless you put that part in a group that you reference in the replacement text, the end of the line will not be in the output.

As a further illustration of groups and backreferences, this command swaps the part before the match and the part after the match.

sed -n -e 's/^\(.*\)\(stalled: \)\(.*\)$/\3\2\1/p'

You can use grep and find combination as below :

find /{foo,fooLib,fooHandler,fooGizmo} -type f -exec grep -l "test" {} \;

Or you can use

find /foo* -type f -exec grep -l "test" {} \;

Or using grep only

grep -R "test" /foo*