regular expression matching a string that is followed with another string without capturing the latter

Question

Is there an ability to make a lookahead assertion non-capturing? Things like bar(?:!foo) and bar(?!:foo) do not work (Python).

Lookaheads *are* non-capturing. Are you perhaps looking for *negative* lookahead? That's just `(?!foo)`. [ref](http://www.regular-expressions.info/lookaround.html) — Alan Moore, Mar 25 '12 at 22:58

score 3 · Accepted Answer · answered Mar 29 '12 at 10:15

3

If you do bar(?=ber) on "barber", "bar" is matched, but "ber" is not captured.

answered Mar 29 '12 at 10:15

zx81

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score 1 · Answer 2 · answered May 23 '12 at 17:59

You didn't respond to Alan's question, but I'll assume that he's correct and you're interested in a negative lookahead assertion. IOW - match 'bar' but NOT 'barfoo'. In that case, you can construct your regex as follows:

myregex =  re.compile('bar(?!foo)')

for example, from the python console:

>>> import re
>>> myregex =  re.compile('bar(?!foo)')
>>> m = myregex.search('barfoo')
>>> print m.group(0)                <=== Error here because match failed
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
AttributeError: 'NoneType' object has no attribute 'group'
>>> m = myregex.search('bar')    
>>> print m.group(0)                <==== SUCCESS!
bar

regular expression matching a string that is followed with another string without capturing the latter

2 Answers2