Omer wrote:
Bob, I tried your way.
>>> import re
>>> urlMask = r"http://[\w\Q./\?=\R]+()?"
>>> text=u"Not working
examplehttp://this.is.a/url?header=nullAnd another
linehttp://and.another.url";
>>> re.findall(urlMask,text)
[u'', u'']
Oops I failed to notice you were using findall. Kent expl
On Mon, Jan 5, 2009 at 11:16 AM, Omer wrote:
> Bob, I tried your way.
>
import re
urlMask = r"http://[\w\Q./\?=\R]+()?"
text=u"Not working examplehttp://this.is.a/url?header=nullAnd
another linehttp://and.another.url";
re.findall(urlMask,text)
> [u'', u'']
>
> spir, I did
Bob, I tried your way.
>>> import re
>>> urlMask = r"http://[\w\Q./\?=\R]+()?"
>>> text=u"Not working examplehttp://this.is.a/url?header=nullAnd
another linehttp://and.another.url";
>>> re.findall(urlMask,text)
[u'', u'']
spir, I did understand it. What I'm not understanding is why isn't this
wor
On Sun, 04 Jan 2009 14:09:53 -0500
bob gailer wrote:
> Omer wrote:
> > I'm sorry, burrowed into the reference until my eyes bled.
> >
> > What I want is to have a regular expression with an optional ending of
> > ""
> >
> > (For those interested,
> > urlMask = r"http://[\w\Q./\?=\R]+";
> > is th
Omer wrote:
I'm sorry, burrowed into the reference until my eyes bled.
What I want is to have a regular expression with an optional ending of
""
(For those interested,
urlMask = r"http://[\w\Q./\?=\R]+";
is ther version w/o the optional ending.)
I can't seem to make a string optional- only
I'm sorry, burrowed into the reference until my eyes bled.
What I want is to have a regular expression with an optional ending of
""
(For those interested,
urlMask = r"http://[\w\Q./\?=\R]+";
is ther version w/o the optional ending.)
I can't seem to make a string optional- only a single charact
