urllib2 file name

Question

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

Answer 1

import os,urllib2
resp = urllib2.urlopen('http://www.example.com/index.html')
my_url = resp.geturl()

os.path.split(my_url)[1]

# 'index.html'

This is not openfile, but maybe still helps :)