How to get everything after last slash in a URL?

Question

How can I extract whatever follows the last slash in a URL in Python? For example, these URLs should return the following:

URL: http://www.test.com/TEST1
returns: TEST1

URL: http://www.test.com/page/TEST2
returns: TEST2

URL: http://www.test.com/page/page/12345
returns: 12345

I've tried urlparse, but that gives me the full path filename, such as page/page/12345.

Answer 1

partition and rpartition are also handy for such things:

url.rpartition('/')[2]

Answer 2

urlparse is fine to use if you want to (say, to get rid of any query string parameters).

import urllib.parse

urls = [
    'http://www.test.com/TEST1',
    'http://www.test.com/page/TEST2',
    'http://www.test.com/page/page/12345',
    'http://www.test.com/page/page/12345?abc=123'
]

for i in urls:
    url_parts = urllib.parse.urlparse(i)
    path_parts = url_parts[2].rpartition('/')
    print('URL: {}\nreturns: {}\n'.format(i, path_parts[2]))

Output:

URL: http://www.test.com/TEST1
returns: TEST1

URL: http://www.test.com/page/TEST2
returns: TEST2

URL: http://www.test.com/page/page/12345
returns: 12345

URL: http://www.test.com/page/page/12345?abc=123
returns: 12345

Answer 3

You don't need fancy things, just see the string methods in the standard library and you can easily split your url between 'filename' part and the rest:

url.rsplit('/', 1)

So you can get the part you're interested in simply with:

url.rsplit('/', 1)[-1]

Answer 4

You cand do like this:

head, tail = os.path.split(url)

Where tail will be your file name.

Answer 5

url ='http://www.test.com/page/TEST2'.split('/')[4]
print url

Output: TEST2.