urllib2.URLError: <urlopen error unknown url type: http

Chris Rebert clp2 at rebertia.com
Wed Mar 11 09:08:48 CET 2009


On Tue, Mar 10, 2009 at 11:34 PM, Coonay <flankw at gmail.com> wrote:
> i use python2.6
>
>
> File "C:\PROGRA~1\Python26\lib\urllib2.py", line 383, in open
>    response = self._open(req, data)
>  File "C:\PROGRA~1\Python26\lib\urllib2.py", line 401, in _open
>    '_open', req)
>  File "C:\PROGRA~1\Python26\lib\urllib2.py", line 361, in
> _call_chain
>    result = func(*args)
>  File "C:\PROGRA~1\Python26\lib\urllib2.py", line 690, in <lambda>
>    meth(r, proxy, type))
>  File "C:\PROGRA~1\Python26\lib\urllib2.py", line 713, in proxy_open
>    return self.parent.open(req)
>  File "C:\PROGRA~1\Python26\lib\urllib2.py", line 383, in open
>    response = self._open(req, data)
>  File "C:\PROGRA~1\Python26\lib\urllib2.py", line 406, in _open
>    'unknown_open', req)
>  File "C:\PROGRA~1\Python26\lib\urllib2.py", line 361, in
> _call_chain
>    result = func(*args)
>  File "C:\PROGRA~1\Python26\lib\urllib2.py", line 1163, in
> unknown_open
>    raise URLError('unknown url type: %s' % type)
> urllib2.URLError: <urlopen error unknown url type: "http>

A. Please include the *entire* traceback, in the future
B. Please include a snippet of the code you're using, in the future
C. A bit less curt of a missive would also be appreciated

That all said, judging by the error message, it appears that the URL
you're trying to open has a double-quote (") as its first character,
which is obviously not permissible.
I can't be completely sure however, since you neglected to include the
entire traceback or any of your actual code.

Cheers,
Chris

-- 
I have a blog:
http://blog.rebertia.com



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