Python split url to find image name and extension

Question

I am looking for a way to extract a filename and extension from a particular url using Python

lets say a URL looks as follows

picture_page = "http://distilleryimage2.instagram.com/da4ca3509a7b11e19e4a12313813ffc0_7.jpg"

How would I go about getting the following.

filename = "da4ca3509a7b11e19e4a12313813ffc0_7"
file_ext = ".jpg"

Answer 1

# Here's your link:
picture_page = "http://distilleryimage2.instagram.com/da4ca3509a7b11e19e4a12313813ffc0_7.jpg"

#Here's your filename and ext:
filename, ext = (picture_page.split('/')[-1].split('.'))

When you do picture_page.split('/'), it will return a list of strings from your url split by a /. If you know python list indexing well, you'd know that -1 will give you the last element or the first element from the end of the list. In your case, it will be the filename: da4ca3509a7b11e19e4a12313813ffc0_7.jpg

Splitting that by delimeter ., you get two values: da4ca3509a7b11e19e4a12313813ffc0_7 and jpg, as expected, because they are separated by a period which you used as a delimeter in your split() call.

Now, since the last split returns two values in the resulting list, you can tuplify it. Hence, basically, the result would be like:

filename,ext = ('da4ca3509a7b11e19e4a12313813ffc0_7', 'jpg')

Answer 2

try:
    # Python 3
    from urllib.parse import urlparse
except ImportError:
    # Python 2
    from urlparse import urlparse
from os.path import splitext, basename

picture_page = "http://distilleryimage2.instagram.com/da4ca3509a7b11e19e4a12313813ffc0_7.jpg"
disassembled = urlparse(picture_page)
filename, file_ext = splitext(basename(disassembled.path))

Only downside with this is that your filename will contain a preceding / which you can always remove yourself.