a regual expression problem

Arnaud Delobelle arnodel at googlemail.com
Sun Nov 30 11:19:37 CET 2008


Arnaud Delobelle <arnodel at googlemail.com> writes:

> 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']

I forgot and re solution!

>>> m=re.search(r'[^/?]*(?=\?|$)', url)
>>> m.group()
'aaa.jpg'

Use split('.') as above if you want name and extension separately.

You could also use something like this:

>>> url[url.rindex('/')+1:url.index('?')]

But it won't work so easily if there is no '?' in the string.

-- 
Arnaud



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