a regual expression problem

[Arnaud Delobelle]

lookon <areyoulookon at gmail.com> writes:

> I have a url of image, and I want to get the filename and extension of
> the image. How to write in python?
>
> for example, the url is http://a.b.com/aaa.jpg?version=1.1
>
> how can I get aaa and jpg by python?

Without res:

>>> url=" http://a.b.com/aaa.jpg?version=1.1"
>>> url.rsplit('/', 1)[1].split('?')[0].split('.')
['aaa', 'jpg']

-- 
Arnaud