Re: using openurl to log into Yahoo services
- From: Dan Sommers <me@xxxxxxxxxxx>
- Date: Wed, 16 Nov 2005 06:29:40 -0500
On Tue, 15 Nov 2005 16:08:06 -0000,
"joe_public34" <joe_public34@xxxxxxxxx> wrote:
> Hello all,
> I'm trying to write a script to log into Yahoo! (mail, groups, etc),
> but when I pass the full URL with all form elements via Python, I get
> a resutling 400 - Bad Request page. When I just plop the full URL
> into a browser, it works perfectly. Is there something I need to
> specify in python to submit the correct info (headers, user agent,
> etc) to get the response I want? Here is the code I have that returns
> the Bad Request:
> import urllib, win32gui, win32clipboard, win32con, os, getpass, re
> yid = raw_input("Yahoo! ID: ")
> pw = getpass.getpass(prompt = 'Yahoo password: ')
> url =
> "https://login.yahoo.com/config/login?.tries=1&.src=ygrp&.intl=us&.v=0&.challenge=U4VY1YGqdPf8z3SaVccJdhV63YCw&.chkP=Y&.done=http://groups.yahoo.com&login="+yid+"&passwd="+pw+"&.persistent=y&.save=Sign
> In"
> temp = urllib.urlopen(url)
> grp_list_source = temp.read()
> Any thoughts or suggestions? Thanks.
It is possible that the "challenge" field is created by yahoo's server
based on the request that caused the server to serve the login page;
perhaps it contains a hash of the user agent that sent the request.
When you reply to that challenge at a much later time, or with a
different user agent, or with something else that's different from the
original request, yahoo's server thinks you're trying to break into
yahoo.
Regards,
Dan
--
Dan Sommers
<http://www.tombstonezero.net/dan/>
.
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- From: joe_public34
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