Hi rishi,
Thanks for ur reply,
yes i set the following enviroment variables (FC6 platform)
http_proxy,http_user,http_password
But i get the same error; Can u tell me which other variables i need to set
or am i going wrong in the syntax of these
variables?
Regards
sv
On Fri, Dec 5, 2008 at 11:57 AM, rishi pathak <[EMAIL PROTECTED]>wrote:
> Are you sitting behind a proxy. If so then you have to set proxy for http
>
> On Fri, Dec 5, 2008 at 11:47 AM, svalbard colaco <[EMAIL PROTECTED]
> > wrote:
>
>> Hi all
>>
>> I have written a small code snippet to open a URL using urllib2 to open a
>> web page , my python version is 2.4 but i get an urlopen error called
>> connection timed out
>>
>> The following is the code snippet
>>
>> *import urllib2
>>
>> f = urllib2.urlopen('http://www.google.com/')
>> print f.read(100)*
>>
>>
>> where as the same url http://www.google.com/ works through my browser.
>>
>> The following is the back trace :
>>
>> File "test_url.py", line 3, in ?
>> f = urllib2.urlopen('http://www.google.com/')
>> File "/usr/lib/python2.4/urllib2.py", line 130, in urlopen
>> return _opener.open(url, data)
>> File "/usr/lib/python2.4/urllib2.py", line 358, in open
>> response = self._open(req, data)
>> File "/usr/lib/python2.4/urllib2.py", line 376, in _open
>> '_open', req)
>> File "/usr/lib/python2.4/urllib2.py", line 337, in _call_chain
>> result = func(*args)
>> File "/usr/lib/python2.4/urllib2.py", line 1021, in http_open
>> return self.do_open(httplib.HTTPConnection, req)
>> File "/usr/lib/python2.4/urllib2.py", line 996, in do_open
>> raise URLError(err)
>> *urllib2.URLError: <urlopen error (110, 'Connection timed out*')>
>>
>>
>> Any pointers in this regard will be of great help.
>>
>> Thanking you'll in advance.
>>
>> Regards,
>> sv
>>
>>
>>
>> --
>> http://mail.python.org/mailman/listinfo/python-list
>>
>>
>
>
> --
> Regards--
> Rishi Pathak
> Pune-Maharastra
>
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