Can not get urllib.urlopen to work
Steve Holden
steve at holdenweb.com
Thu Oct 28 00:45:33 EDT 2004
Pater Maximus wrote:
> I am trying to implement the recipe listed at
> http://aspn.activestate.com/ASPN/Cookbook/Python/Recipe/211886
>
> However, I can not get to first base. When I try to run
>
> import urllib
> fo=urllib.urlopen("http://www.dictionary.com/")
> page = fo.read()
>
> I get:
>
> Traceback (most recent call last):
> File "C:/Program Files/Python/Lib/idlelib/testurl", line 2, in -toplevel-
> fo=urllib.urlopen("http://www.dictionary.com/")
> File "C:\PROGRA~1\PYTHON\lib\urllib.py", line 76, in urlopen
> return opener.open(url)
> File "C:\PROGRA~1\PYTHON\lib\urllib.py", line 181, in open
> return getattr(self, name)(url)
> File "C:\PROGRA~1\PYTHON\lib\urllib.py", line 297, in open_http
> h.endheaders()
> File "C:\PROGRA~1\PYTHON\lib\httplib.py", line 712, in endheaders
> self._send_output()
> File "C:\PROGRA~1\PYTHON\lib\httplib.py", line 597, in _send_output
> self.send(msg)
> File "C:\PROGRA~1\PYTHON\lib\httplib.py", line 564, in send
> self.connect()
> File "C:\PROGRA~1\PYTHON\lib\httplib.py", line 548, in connect
> raise socket.error, msg
> IOError: [Errno socket error] (10061, 'Connection refused')
>
>
Suspect the action of a firewall or just assume you were unlucky and the
server was down when you hit it. I had no problem just now:
$ python
Python 2.3.4 (#1, Jun 13 2004, 11:21:03)
[GCC 3.3.1 (cygming special)] on cygwin
Type "help", "copyright", "credits" or "license" for more information.
>>> import urllib
>>> fo=urllib.urlopen("http://www.dictionary.com/")
>>> page = fo.read()
>>>
regards
Steve
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