Re: a bug in urllib?

From: GMTaglia (liquid_at_kuht.it)
Date: 07/04/04 

Date: 4 Jul 2004 08:38:56 -0700

Haim Ashkenazi <haim@babysnakes.org> wrote in message news:<mailman.387.1088880858.27577.python-list@python.org>...
> Hi
>
> I'm writing a script that uses urllib on win98. until now I used python
> 2.3.x (x < 4) and it worked ok. I re-installed windows and installed
> python 2.3.4 and now I get an error when trying to open a url "no host
> given":
>
> Python 2.3.4 (#53, May 25 2004, 21:17:02) [MSC v.1200 32 bit (Intel)] on win32
>
> Type "copyright", "credits" or "license()" for more information.
>
>
>
> ****************************************************************
>
> Personal firewall software may warn about the connection IDLE
>
> makes to its subprocess using this computer's internal loopback
>
> interface. This connection is not visible on any external
>
> interface and no data is sent to or received from the Internet.
>
> ****************************************************************
>
>
>
> IDLE 1.0.3
>
> >>> import urllib
>
> >>> f = urllib.URLopener()
>
> >>> x = f.open('http://www.python.org')
>
>
>
> Traceback (most recent call last):
>
> File "<pyshell#2>", line 1, in -toplevel-
>
> x = f.open('http://www.python.org')
>
> File "C:\PYTHON23\Lib\urllib.py", line 181, in open
>
> return getattr(self, name)(url)
>
> File "C:\PYTHON23\Lib\urllib.py", line 281, in open_http
>
> if not host: raise IOError, ('http error', 'no host given')
>
> IOError: [Errno http error] no host given
>
> >>>
>
> --------------------------------------------
>
> I don't think it's a network error because I can browse to this site
> without a problem. does anyone have any idea? is it a bug? can anyone tell
> me where I can download an earlier version (not activestate, it's not
> running well on my win98)?
>
> Bye

I've just tried also in windows but no errors, same version as you

Python 2.3.4 (#53, May 25 2004, 21:17:02) [MSC v.1200 32 bit (Intel)] on win32
Type "copyright", "credits" or "license()" for more information.

>>> import urllib
>>> f = urllib.URLopener()
>>> x = f.open('http://www.python.org')
>>>

Maybe you can reconsider a network error?
ciao
Mario

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