urllib2 file name

If I open a file using urllib2, like so:

remotefile = urllib2.urlopen('http://example.com/somefile.zip')

Is there an easy way to get the file name other then parsing the original URL?

EDIT: changed openfile to urlopen... not sure how that happened.

EDIT2: I ended up using:

filename = url.split('/')[-1].split('#')[0].split('?')[0]

Unless I'm mistaken, this should strip out all potential queries as well.

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('/')[-1]
'somefile.zip'
>>> 'http://example.com/somedir/somefile.zip'.split('/')[-1]
'somefile.zip'

If you only want the file name itself, assuming that there's no query variables at the end like http://example.com/somedir/somefile.zip?foo=bar then you can use os.path.basename for this:

[user@host]$ python
Python 2.5.1 (r251:54869, Apr 18 2007, 22:08:04) 
Type "help", "copyright", "credits" or "license" for more information.
>>> import os
>>> os.path.basename("http://example.com/somefile.zip")
'somefile.zip'
>>> os.path.basename("http://example.com/somedir/somefile.zip")
'somefile.zip'
>>> os.path.basename("http://example.com/somedir/somefile.zip?foo=bar")
'somefile.zip?foo=bar'

Some other posters mentioned using urlparse, which will work, but you'd still need to strip the leading directory from the file name. If you use os.path.basename() then you don't have to worry about that, since it returns only the final part of the URL or file path.

I think that "the file name" isn't a very well defined concept when it comes to http transfers. The server might (but is not required to) provide one as "content-disposition" header, you can try to get that with remotefile.headers['Content-Disposition']. If this fails, you probably have to parse the URI yourself.

Just saw this I normally do..

filename = url.split("?")[0].split("/")[-1]

Using urlsplit is the safest option:

url = 'http://example.com/somefile.zip'
urlparse.urlsplit(url).path.split('/')[-1]