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
.
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]
One more (idio(ma)tic) way:
URL.split("/")[-1]
rsplit
should be up to the task:
In [1]: 'http://www.test.com/page/TEST2'.rsplit('/', 1)[1]
Out[1]: 'TEST2'
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
You cand do like this:
head, tail = os.path.split(url)
Where tail will be your file name.
extracted_url = url[url.rfind("/")+1:];
Here's a more general, regex way of doing this:
re.sub(r'^.+/([^/]+)$', r'\1', url)
partition
and rpartition
are also handy for such things:
url.rpartition('/')[2]
Split the url and pop the last element
url.split('/').pop()
os.path.basename(os.path.normpath('/folderA/folderB/folderC/folderD/'))
folderD
url ='http://www.test.com/page/TEST2'.split('/')[4]
print url
Output: TEST2
.