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
ret         


        

Answer 1

rsplit should be up to the task:

In [1]: 'http://www.test.com/page/TEST2'.rsplit('/', 1)[1]
Out[1]: 'TEST2'

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

Here's a more general, regex way of doing this:

    re.sub(r'^.+/([^/]+)$', r'\1', url)

Answer 4

One more (idio(ma)tic) way:

URL.split("/")[-1]

Answer 5

os.path.basename(os.path.normpath('/folderA/folderB/folderC/folderD/'))

>>> folderD

Answer 6

You can do like this:

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

Where tail will be your file name.