Grep in bash with regex

Question

I am getting the following output from a bash script:

INFOPLIST_FILE = MajorDomo/MajorDomo-Info.plist

and I would like to get only the path(MajorDomo/MajorDomo-Info.plist) using grep. In other words, everything after the equals sign. Any ideas of how to do this?

Answer 1

This job suites more to awk:

s='INFOPLIST_FILE = MajorDomo/MajorDomo-Info.plist'
awk -F' *= *' '{print $2}' <<< "$s"
MajorDomo/MajorDomo-Info.plist

If you really want grep then use grep -P:

grep -oP ' = \K.+' <<< "$s"
MajorDomo/MajorDomo-Info.plist

Answer 2

Not exactly what you were asking, but

echo "INFOPLIST_FILE = MajorDomo/MajorDomo-Info.plist" | sed 's/.*= \(.*\)$/\1/'

will do what you want.