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
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