a regual expression problem

[lookon]

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?

 

[Arnaud Delobelle]

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 Delobelle]

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.

 

[Jon Clements]

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 ishttp://a.b.com/aaa.jpg?version=1.1

how can I get aaa and jpg by python?

Something like...

from urlparse import urlsplit
from os.path import splitext
splitext(urlsplit('http://a.b.com/aaa.jpg?version=1.1').path[1:])

('aaa', '.jpg')

....and then it should be fairly flexible instead of having to change
splitting etc..

hth

Jon.

 

[Roy Smith]

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?

Despite the subject line, this is *not* a regex problem. The right tool is
Python's built-in urlparse module, which understands the full syntax of
urls. Trying to roll your own from scratch using split() or regex is the
wrong solution.