a bug in urllib?

[GMTaglia]

Haim Ashkenazi <haim at babysnakes.org> wrote in message news:<mailman.387.1088880858.27577.python-list at 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