Bugs item #1368368, was opened at 2005-11-28 17:39
Message generated for change (Tracker Item Submitted) made by Item Submitter
You can respond by visiting:
https://sourceforge.net/tracker/?func=detail&atid=105470&aid=1368368&group_id=5470
Please note that this message will contain a full copy of the comment thread,
including the initial issue submission, for this request,
not just the latest update.
Category: None
Group: Python 2.5
Status: Open
Resolution: None
Priority: 5
Submitted By: Björn Lindqvist (sonderblade)
Assigned to: Nobody/Anonymous (nobody)
Summary: prompt_user_passwd() in FancyURLopener
Initial Comment:
Currently, urllib.urlopen() "kind of" handles HTTP
authentication. You simply write something like this:
urllib.urlopen("http://foo:[EMAIL PROTECTED]")
And it works like a charm. EXCEPT when the
username/password is wrong. Then you get the
FancyURLopener.prompt_user_passwd()-prompt asking you
for your username and password. I think that behaviour
is extremely illogical and annoying because it forces
you to add more code just to circumvent the odd default
behaviour.
I would be MUCH happer if, instead of the prompt, you
get an exception or the 401 Unauthorized error page
returned.
----------------------------------------------------------------------
You can respond by visiting:
https://sourceforge.net/tracker/?func=detail&atid=105470&aid=1368368&group_id=5470
_______________________________________________
Python-bugs-list mailing list
Unsubscribe:
http://mail.python.org/mailman/options/python-bugs-list/archive%40mail-archive.com
