is there a possibillity to get the filename
e.g. xyz.com/blafoo/showall.html if you work with urllib or httplib?
so that i can save the file under the filename on the server?
if you go to sites like
xyz.com/blafoo/ you cant see the filename.
Thank you
To get filename from response http headers:
import cgi response = urllib2.urlopen(URL) _, params = cgi.parse_header(response.headers.get('Content-Disposition', '')) filename = params['filename'] To get filename from the URL:
import posixpath import urlparse path = urlparse.urlsplit(URL).path filename = posixpath.basename(path) 
posixpath is the correct module here. Moreover it would be a mistake to use os.path here. If you can't figure out "why", ask, I'll elaborate.
'/' in their path segment. os.path on Windows may use '\\' that is not appropriate for urls as pathname separator. posixpath uses '/'.Use urllib.request.Request:
import urllib req = urllib.request.Request(url, method='HEAD') r = urllib.request.urlopen(req) print(r.info().get_filename()) Example :
In[1]: urllib.request.urlopen(urllib.request.Request('https://httpbin.org/response-headers?content-disposition=%20attachment%3Bfilename%3D%22example.csv%22', method='HEAD')).info().get_filename() Out[1]: 'example.csv' Does not make much sense what you are asking. The only thing that you have is the URL. Either extract the last part from the URL or you may check the HTTP response for something like
content-disposition: attachment;filename="foo.bar" This header can be set by the server to indicate that the filename is foo.bar. This is usually used for file downloads or something similar.
I searched for you question on google and I saw that it was answered in stackoverflow before I believe.
Try looking at this post:
Using urllib2 in Python. How do I get the name of the file I am downloading?
The filename is usually included by the server through the content-disposition header:
content-disposition: attachment; filename=foo.pdfYou have access to the headers through
result = urllib2.urlopen(...) result.info() <- contains the headers i>>> import urllib2 ur>>> result = urllib2.urlopen('http://zopyx.com') >>> print result <addinfourl at 4302289808 whose fp = <socket._fileobject object at 0x1006dd5d0>> >>> result.info() <httplib.HTTPMessage instance at 0x1006fbab8> >>> result.info().headers ['Date: Mon, 04 Apr 2011 02:08:28 GMT\r\n', 'Server: Zope/(unreleased version, python 2.4.6, linux2) ZServer/1.1Plone/3.3.4\r\n', 'Content-Length: 15321\r\n', 'Content-Type: text/html; charset=utf-8\r\n', 'Via: 1.1 www.zopyx.com\r\n', 'Cache-Control: max-age=3600\r\n', 'Expires: Mon, 04 Apr 2011 03:08:28 GMT\r\n', 'Connection: close\r\n']
See
