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 <[EMAIL PROTECTED]>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 <[EMAIL PROTECTED] > > 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 <[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 >>> >> >> > > > -- > Regards-- > Rishi Pathak > Pune-Maharastra >
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