> -----Original Message-----
> From: [EMAIL PROTECTED] [mailto:python-
> [EMAIL PROTECTED] On Behalf Of Reedick, Andrew
> Sent: Tuesday, July 01, 2008 10:07 AM
> To: [EMAIL PROTECTED]; python-list@python.org
> Subject: RE: How make regex that means "contains regex#1 but NOT
> regex#2" ??
>
> Match 'foo.*bar', except when 'not' appears between foo and bar.
>
>
> import re
>
> s = 'fooAAABBBbar'
> print "Should match:", s
> m = re.match(r'(foo(.(?!not))*bar)', s);
> if m:
> print m.groups()
>
> print
>
> s = 'fooAAAnotBBBbar'
> print "Should not match:", s
> m = re.match(r'(foo(.(?!not))*bar)', s);
> if m:
> print m.groups()
>
>
> == Output ==
> Should match: fooAAABBBbar
> ('fooAAABBBbar', 'B')
>
> Should not match: fooAAAnotBBBbar
>
Fixed a bug with 'foonotbar'. Conceptually it breaks down into:
First_half_of_Regex#1(not
Regex#2)(any_char_Not_followed_by_Regex#2)*Second_half_of_Regex#1
However, if possible, I would make it a two pass regex. Match on
Regex#1, throw away any matches that then match on Regex#2. A two pass
is faster and easier to code and understand. Easy to understand == less
chance of a bug. If you're worried about performance, then a) a
complicated regex may or may not be faster than two simple regexes, and
b) if you're passing that much data through a regex, you're probably I/O
bound anyway.
import re
ss = ('foobar', 'fooAAABBBbar', 'fooAAAnotBBBbar', 'fooAAAnotbar',
'foonotBBBbar', 'foonotbar')
for s in ss:
print s,
m = re.match(r'(foo(?!not)(?:.(?!not))*bar)', s);
if m:
print m.groups()
else:
print
== output ==
foobar ('foobar',)
fooAAABBBbar ('fooAAABBBbar',)
fooAAAnotBBBbar
fooAAAnotbar
foonotBBBbar
foonotbar
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