RFC 2368 <http://www.ietf.org/rfc/rfc2368.txt> specifies mailto: URLs as
having the following syntax:

     mailtoURL  =  "mailto:" [ to ] [ headers ]
     to         =  #mailbox
     headers    =  "?" header *( "&" header )
     header     =  hname "=" hvalue
     hname      =  *urlc
     hvalue     =  *urlc

The header fields in these URLs are roughly analogous to query
parameters in other URLs, but urlparse treats them as part of the path
(along with the email address):

>>> import urlparse
>>> urlparse.urlparse("mailto:foo@example.com?subject=hi")
ParseResult(scheme='mailto', netloc='',
path='foo@example.com?subject=hi', params='', query='', fragment='')

It should treat them as query parameters instead, which would not only
make it easier to access them but would also make it easier to access
the email address, since one would no longer have to parse headers, if
any, out of the path first.

Adding the 'mailto' scheme to the urllib.parse.uses_query list changes the behavior as described in msg91249. A patch with a test is attached.

Note that this changes the behavior of urllib.parse.urlparse() on 'mailto:' URLs: with this patch, the function returns the "foo@example.com" in the "netloc" named tuple position instead of in the "path" position.

Which would be a bug.  According to RFC3986 it should be the path.  So the change can't be quite so simple as adding mailto to uses_query.  Since the query component is defined for generic uris according to the RFC, it sounds like urllib's parser needs some enhancement.

Tested on 3.4

urllib.parse.urlparse("mailto:foo@example.com?subject=hi")
ParseResult(scheme='mailto', netloc='', path='foo@example.com', params='', query='subject=hi', fragment='')

Work as expected.

RFC3986 bears the weightage since urlparse aims to satisfy that. As per that, mailto should return the url as path, and current urlparse behaves properly.

$ ./python.exe
Python 3.4.0a0 (default:50164abbfc98+, Apr  8 2013, 22:19:34)
[GCC 4.2.1 (Apple Inc. build 5666) (dot 3)] on darwin
Type "help", "copyright", "credits" or "license" for more information.
>>> import urllib.parse
>>> urllib.parse.urlparse("mailto:foo@example.com?subject=hi")
ParseResult(scheme='mailto', netloc='', path='foo@example.com', params='', query='subject=hi', fragment='')
>>>

So, I am closing this bug as won't fix.