Amrith Kumar added the comment:
As requested ...
>>> urlparse.urlparse('http://www.google.com:/abc')
ParseResult(scheme='http', netloc='www.google.com:', path='/abc', params='',
query='', fragment='')
I submit to you that this should generate an error.
----------
_______________________________________
Python tracker <[email protected]>
<http://bugs.python.org/issue28841>
_______________________________________
_______________________________________________
Python-bugs-list mailing list
Unsubscribe:
https://mail.python.org/mailman/options/python-bugs-list/archive%40mail-archive.com
