CONNECTION TIMED OUT ERROR using urllib2

[svalbard colaco]

Hi

I set my http_proxy and now
i get the following error

*urllib2.HTTPError: HTTP Error 403: Forbidden ( The ISA Server denied the
specified Uniform Resource Locator (URL).  *

what other variables have to be set ?

Regards,
sv

On Fri, Dec 5, 2008 at 12:47 PM, rishi pathak <mailmaverick666 at gmail.com>wrote:

> Before executing script do
> export http_proxy=http://<your proxy server address>:<port>/
>
>
>
> On Fri, Dec 5, 2008 at 12:06 PM, svalbard colaco <svalbardcolaco at gmail.com
> > wrote:
>
>> 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 <mailmaverick666 at gmail.com>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 <
>>> svalbardcolaco at gmail.com> 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
>>>
>>
>>
>
>
> --
> Regards--
> Rishi Pathak
> Pune-Maharastra
>
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