New submission from Jonathan Windle <jonathanwin...@gmail.com>: Example Code: import re re.findall(r"(?![a-z0-9])0(?![a-z0-9])", "a0a 0 0 b0b")
The above code returns an empty list. I expect to get ['0', '0'] returned. If I remove "(?![a-z0-9])" from the beginning of the expression string findall returns values as expected. ---------- components: Regular Expressions messages: 92046 nosy: jwindle severity: normal status: open title: When Beginning Expression with Lookahead Assertion I get no Matches type: behavior versions: Python 2.6 _______________________________________ Python tracker <rep...@bugs.python.org> <http://bugs.python.org/issue6797> _______________________________________ _______________________________________________ Python-bugs-list mailing list Unsubscribe: http://mail.python.org/mailman/options/python-bugs-list/archive%40mail-archive.com
