Bugs item #1722348, was opened at 2007-05-20 18:35
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Category: Python Library
Group: None
Status: Open
Resolution: None
Priority: 5
Private: No
Submitted By: Thomas Folz-Donahue (eigenlambda)
Assigned to: Nobody/Anonymous (nobody)
Summary: urlparse.urlunparse forms file urls incorrectly
Initial Comment:
This is a conversation with the current Python interpreter.
>>> import urlparse
>>> urlparse.urlparse(urlparse.urlunparse(urlparse.urlparse("file:////usr/bin/python")))
('file', 'usr', '/bin/python', '', '', '')
As you can see, the results are incorrect. The problem is in the urlunsplit
function:
def urlunsplit((scheme, netloc, url, query, fragment)):
if netloc or (scheme and scheme in uses_netloc and url[:2] != '//'):
if url and url[:1] != '/': url = '/' + url
url = '//' + (netloc or '') + url
if scheme:
url = scheme + ':' + url
if query:
url = url + '?' + query
if fragment:
url = url + '#' + fragment
return url
RFC 1808 (see http://www.ietf.org/rfc/rfc1808.txt ) specifies that a URL shall
have the following syntax:
<scheme>://<net_loc>/<path>;<params>?<query>#<fragment>
The problem with the current version of urlunsplit is that it tests if there
are already two slashes before the 'url' section before outputting a URL. This
is incorrect because (1) RFC 1808 clearly specifies at least three slashes
between the end of the scheme portion and the beginning of the path portion and
(2) this method will strip the first few slashes from an arbitrary path
portion, which may require those slashes. Removing that url[:2] != '//' causes
urlunsplit to behave correctly when dealing with urls like
file:////usr/bin/python .
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