Regular expressions using Python3 with greedy modifier not working

0
votes
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/
1
python-3.xpython-re
Did you just want to match any instance that's not at the beginning of the string? If so, get rid of the "?" and see Regex: match pattern as long as it's not in the beginning – SuperStormer
It doesn't work. When I do this, I get the entire string. – Gary
Did you follow the linked post? – SuperStormer
Yes, thank you for your help, but that post isn't the solution I need. – Gary
I need the last match in a string, even if it's at the beginning (and consequently the last one). e.g. facebook. com/hello/ – Gary

1 Answers

0
votes

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