Bob, I tried your way. >>> import re >>> urlMask = r"http://[\w\Q./\?=\R]+(<br>)?" >>> text=u"Not working example<br>http://this.is.a/url?header=null<br>And another line<br>http://and.another.url"; >>> re.findall(urlMask,text) [u'<br>', u'']
spir, I did understand it. What I'm not understanding is why isn't this working. (Whereas, >>> OldurlMask = r"http://[\w\Q./\?=\R]+"; #Not f-ing working. >>> re.findall(OldurlMask,text) ['http://this.is.a/url?header=null', 'http://and.another.url'] does work. Which is what had me frowning. Also, this ugly url mask is working: >>> UglyUrlMask = r"(http://[\w\Q./\?=\R]+<br>|http://[\w\Q./\?=\R]+)" >>> re.findall(UglyUrlMask,text) ['http://this.is.a/url?header=null<br>', 'http://and.another.url'] Anyone?) On Mon, Jan 5, 2009 at 12:08 AM, spir <denis.s...@free.fr> wrote: > On Sun, 04 Jan 2009 14:09:53 -0500 > bob gailer <bgai...@gmail.com> 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 > > > "<br>" > > > > > > (For those interested, > > > urlMask = r"http://[\w\Q./\?=\R]+"; > > > is ther version w/o the optional <br> ending.) > > > > > > I can't seem to make a string optional- only a single character via > > > []s. I for some reason thuoght it'll be ()s, but no help there- it > > > just returns only the <br>. Anybody? > > > > > urlMask = r"http://[\w\Q./\?=\R]+(<br>)?" > > > > From the docs: ? Causes the resulting RE to match 0 or 1 repetitions of > > the preceding RE. ab? will match either 'a' or 'ab'. > > > > > > Maybe Omer had not noted that a sub-expression can be grouped in () so that > an operator (?+*) applies on the whole group. > Denis > > ------ > la vida e estranya > _______________________________________________ > Tutor maillist - Tutor@python.org > http://mail.python.org/mailman/listinfo/tutor >
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