Remove Part Of String Before The Last Forward Slash

The program I am currently working on retrieves URLs from a website and puts them into a list. What I want to get is the last section of the URL. So, if the first element in my lis

Solution 1:

Have a look at str.rsplit.

>>> s = 'https://docs.python.org/3.4/tutorial/interpreter.html'>>> s.rsplit('/',1)
['https://docs.python.org/3.4/tutorial', 'interpreter.html']
>>> s.rsplit('/',1)[1]
'interpreter.html'

And to use RegEx

>>> re.search(r'(.*)/(.*)',s).group(2)
'interpreter.html'

Then match the 2nd group which lies between the last / and the end of String. This is a greedy usage of the greedy technique in RegEx.

Debuggex Demo

Small Note - The problem with link.rpartition('//')[-1] in your code is that you are trying to match // and not /. So remove the extra / as in link.rpartition('/')[-1].

Solution 2:

That doesn't need regex.

import osfor link in link_list:
    file_names.append(os.path.basename(link))

Solution 3:

You can use rpartition():

>>> s = 'https://docs.python.org/3.4/tutorial/interpreter.html'>>> s.rpartition('/')
('https://docs.python.org/3.4/tutorial', '/', 'interpreter.html')

And take the last part of the 3 element tuple that is returned:

>>> s.rpartition('/')[2]
'interpreter.html'

Solution 4:

Just use string.split:

url = "/some/url/with/a/file.html"print url.split("/")[-1]

# Result should be "file.html"

split gives you an array of strings that were separated by "/". The [-1] gives you the last element in the array, which is what you want.

Solution 5:

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

    re.sub(r'^.+/([^/]+)$', r'\1', "http://test.org/3/files/interpreter.html")
    'interpreter.html'