Regular expression only matches start of string

Asked Feb 09 '20 at 22:21

Active Feb 09 '20 at 22:42

Viewed 22 times

I am trying to parse the output of a git command with a python script. To that effect, let's say we want to find all the modified and added files in git status, then we would do something like this:

import glob
import os
import re
import git

def main():
    g = git.cmd.Git('./')

    layout_pattern = '(modified:|new file:).*'
    matches = re.findall(layout_pattern, g.status())
    print(matches)

if __name__ == "__main__":
    main()

However, the output from python is truncated:

['modified:', 'new file:', 'modified:', 'modified:']

There is an explicit .* expression in the pattern, it should match all the way to the end of the line, why isn't it?

edited Feb 09 '20 at 22:42

martineau

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asked Feb 09 '20 at 22:21

Makogan

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`re.findall` returns only capture groups if any. Use a non-capturing group. – Casimir et Hippolyte Feb 09 '20 at 22:24
1

change `(modified:|new file:)` to a non-capturing group `(?:modified:|new file:)` and you will get the entire match – Nick Feb 09 '20 at 22:26
What is a non-capturing group? – Makogan Feb 09 '20 at 22:26
Or use `re.finditer`. See all details at https://stackoverflow.com/a/31915134/3832970 – Wiktor Stribiżew Feb 09 '20 at 22:27
`?:` tells the regex engine not to capture the contents of the group it is in. `re.findall` will only return the contents of capturing groups if there are any in the regex, that is why you are just getting `'modified:'` etc. as your result. – Nick Feb 09 '20 at 22:27
@Nick your solution worked, so if you upgrade it to an answer I'll accept it – Makogan Feb 09 '20 at 22:28
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@Makogan I didn't answer because I was expecting the question to be closed as a dupe... – Nick Feb 09 '20 at 22:29

Regular expression only matches start of string

0 Answers0