Python Split Url To Find Image Name And Extension

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/

Solution 1:

try:
    # Python 3from urllib.parse import urlparse
except ImportError:
    # Python 2from 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.

Solution 2:

Try with urlparse.urlsplit to split url, and then os.path.splitext to retrieve filename and extension (use os.path.basename to keep only the last filename) :

import urlparse
import os.path

picture_page = "http://distilleryimage2.instagram.com/da4ca3509a7b11e19e4a12313813ffc0_7.jpg"printos.path.splitext(os.path.basename(urlparse.urlsplit(picture_page).path))

>>> ('da4ca3509a7b11e19e4a12313813ffc0_7', '.jpg')

Solution 3:

filename = picture_page.split('/')[-1].split('.')[0]
file_ext = '.'+picture_page.split('.')[-1]

Solution 4:

# 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')

Solution 5:

os.path.splitext will help you extract the filename and extension once you have extracted the relevant string from the URL using urlparse:

   fName, ext = os.path.splitext('yourImage.jpg')