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Author sjones
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Date 2003-06-14.02:39:05
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Sorry, previous comment got cut off...

urlparse.urlparse takes a url of the format:
    <scheme>://<netloc>/<path>;<params>?<query>#<fragment>

And returns a 6-tuple of the format:
    (scheme, netloc, path, params, query, fragment).

An example from the library refrence takes:
    urlparse('http://www.cwi.nl:80/%7Eguido/Python.html')

And produces:
    ('http', 'www.cwi.nl:80', '/%7Eguido/Python.html', '',
'', '')

--------------------------------

Note that there isn't a field for the port number in the
6-tuple. Instead, it is included in the netloc. Urlparse
should handle your example as:

>>> urlparse.urlparse('1.2.3.4:80','http') 
('http', '1.2.3.4:80', '', '', '', '')

Instead, it gives the incorrect output as you indicated.
History
Date User Action Args
2007-08-23 14:13:54adminlinkissue754016 messages
2007-08-23 14:13:54admincreate