This PR will make the following changes:

• update the urlsplit and urlunsplit functions of the urllib.parse module to keep the ? delimiter in a URI with an empty query component and keep the # delimiter in a URI with an empty fragment component (currently the delimiters are dropped):

    >>> from urllib.parse import urlsplit, urlunsplit
    >>> urlunsplit(urlsplit('http://example.com/?'))
    'http://example.com/?'  # currently: 'http://example.com/'
    >>> urlunsplit(urlsplit('http://example.com/#'))
    'http://example.com/#'  # currently: 'http://example.com/'
  

  This is required by RFC 3986:

  Normalization should not remove delimiters when their associated component is empty unless licensed to do so by the scheme specification. For example, the URI "http://example.com/?" cannot be assumed to be equivalent to any of the examples above. Likewise, the presence or absence of delimiters within a userinfo subcomponent is usually significant to its interpretation. The fragment component is not subject to any scheme-based normalization; thus, two URIs that differ only by the suffix "#" are considered different regardless of the scheme.

  To do so:

  • the urlsplit function now decodes an absent '' query component as None and an absent '' fragment component as None (e.g., urlsplit('http://example.com/') → ('http', 'example.com', '/', None, None)), and still decodes an empty '?' query component as '' and an empty '#' fragment component as '' (e.g., urlsplit('http://example.com/?#') → ('http', 'example.com', '/', '', ''));
  • the urlunsplit function now encodes a None query component as an absent '' query component and a None fragment component as an absent '' fragment component (e.g., urlunsplit(('http', 'example.com', '/', None, None)) → 'http://example.com/'), and now encodes a '' query component as an empty '?' query component and a '' fragment component as an empty '#' fragment component (e.g., urlunsplit(('http', 'example.com', '/', '', '')) → 'http://example.com/?#');

• add and update the corresponding unit tests in the test.test_urlparse module;

• update a unit test in the test.test_urllib2 module;

• update the urllib.parse documentation accordingly.

https://bugs.python.org/issue37969