Matt Nordhoff wrote: > John Nagle wrote: >> Here's a hostile URL that "urlparse.urlparse" seems to have mis-parsed. >> ==== ...
> > It's breaking on the first slash, which just happens to be very late in > the URL. > >>>> urlparse('http://example.com?blahblah=http://example.net') > ('http', 'example.com?blahblah=http:', '//example.net', '', '', '') That's what it seems to be doing: sa1 = 'http://example.com?blahblah=/foo' sa2 = 'http://example.com?blahblah=foo' print urlparse.urlparse(sa1) ('http', 'example.com?blahblah=', '/foo', '', '', '') # WRONG print urlparse.urlparse(sa2) ('http', 'example.com', '', '', 'blahblah=foo', '') # RIGHT That's wrong. RFC3896 ("Uniform Resource Identifier (URI): Generic Syntax"), page 23 says "The characters slash ("/") and question mark ("?") may represent data within the query component. Beware that some older, erroneous implementations may not handle such data correctly when it is used as the base URI for relative references (Section 5.1), apparently because they fail to distinguish query data from path data when looking for hierarchical separators." So "urlparse" is an "older, erroneous implementation". Looking at the code for "urlparse", it references RFC1808 (1995), which was a long time ago, three revisions back. Here's the bad code: def _splitnetloc(url, start=0): for c in '/?#': # the order is important! delim = url.find(c, start) if delim >= 0: break else: delim = len(url) return url[start:delim], url[delim:] That's just wrong. The domain ends at the first appearance of any character in '/?#', but that code returns the text before the first '/' even if there's an earlier '?'. A URL/URI doesn't have to have a path, even when it has query parameters. This bug is in Python 2.4 and 2.5. I'll file a bug report. John Nagle SiteTruth -- http://mail.python.org/mailman/listinfo/python-list