Did you mean urllib2.urlopen?
You could potentially lift the intended filename if the server was sending a Content-Disposition header by checking remotefile.info()['Content-Disposition']
, but as it is I think you’ll just have to parse the url.
You could use urlparse.urlsplit
, but if you have any URLs like at the second example, you’ll end up having to pull the file name out yourself anyway:
>>> urlparse.urlsplit('http://example.com/somefile.zip')
('http', 'example.com', '/somefile.zip', '', '')
>>> urlparse.urlsplit('http://example.com/somedir/somefile.zip')
('http', 'example.com', '/somedir/somefile.zip', '', '')
Might as well just do this:
>>> 'http://example.com/somefile.zip'.split("https://stackoverflow.com/")[-1]
'somefile.zip'
>>> 'http://example.com/somedir/somefile.zip'.split("https://stackoverflow.com/")[-1]
'somefile.zip'