CODEWITHSUNDEEP
python-3.xpython-re
Gary
Gary

Reputation: 1017

Regular expressions using Python3 with greedy modifier not working

import re
s="facebook.com/https://www.facebook.com/test/"
re.findall("facebook\.com/[^?\"\'>&\*\n\r\\\ <]+?", s)

I only want as a result "facebook.com/test/" ... but I'm getting as a result --

facebook.com/h
facebook.com/t

What's wrong with my RE? I applied the "?" at the end of the expression thinking this would stop greediness, but it's being treated as 0 or 1 expression.

If I remove the "?" I get:

facebook.com/https://www.facebook.com/test/

Upvotes: 1

Views: 28

Answers (1)

blhsing
blhsing

Reputation: 106638

The non-greedy modifier works forwards but not backwards, which means that when the first instance of facebook.com/ matches it will not be discarded unless the rest of the pattern fails to match, even if it's non-greedy.

To match the last instance of facebook.com/ you can use a negative lookahead pattern instead:

facebook\.com/(?!.*facebook\.com/)[^?\"\'>&\*\n\r\\\ <]+

Demo: https://replit.com/@blhsing/WordyAgitatedCallbacks

Upvotes: 1

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