Grep non capturing group and capturing group in the same regex. Print only the capturing group

Question

I have a file that contains something like this:

VERSION = 1.1.1
version = 1.1.1
VERSION = "1.1.1"
VERSION = '1.1.1"

etc

I am trying to grep for the version number, using this:

grep -E "(VERSION|version|Version)[^0-9]*([0-9]+\.[0-9,A-Z,a-z]+\.[0-9,a-z,A-Z]+)" -oP setup.py

This complains that 'conflicting matchers specified'. Presumably because there are two capturing groups. I have tried to make the first group non-capturing: (?:VERSION|version|Version), but this gives the same error. The version with the non-capturing group does work when tested on regex101 (https://regex101.com/r/21Pkp2/1), so I'm not sure why it doesn't work in grep.

Answer 1

You cannot use -E (ERE) and -P (perl mode) options together in gnu grep. If you are using gnu grep then just stick with -P and use this command to get all version numbers:

grep -ioP "version\D*\K(\d+\.[\d,A-Z,a-z]+\.[\d,a-z,A-Z]+)" file

1.1.1
1.1.1
1.1.1
1.1.1

• -i is used for ignore case match.
• \K is PCRE directive to discard previously matched info