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  r81521 | r.david.murray | 2010-05-25 11:32:06 -0400 (Tue, 25 May 2010) | 2 lines

  Issue 8818: urlparse/urlsplit keyword is 'scheme', not 'default_scheme'.
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R. David Murray 2010-05-25 15:49:57 +00:00
parent 5b9de5ccc5
commit 71b2315824
1 changed files with 3 additions and 3 deletions
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Doc/library/urlparse.rst
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@ -36,7 +36,7 @@ following URL schemes: ``file``, ``ftp``, ``gopher``, ``hdl``, ``http``,
The :mod:`urlparse` module defines the following functions:
.. function:: urlparse(urlstring[, default_scheme[, allow_fragments]])
.. function:: urlparse(urlstring[, scheme[, allow_fragments]])
Parse a URL into six components, returning a 6-tuple. This corresponds to the
general structure of a URL: ``scheme://netloc/path;parameters?query#fragment``.
@ -58,7 +58,7 @@ The :mod:`urlparse` module defines the following functions:
>>> o.geturl()
'http://www.cwi.nl:80/%7Eguido/Python.html'
If the *default_scheme* argument is specified, it gives the default addressing
If the *scheme* argument is specified, it gives the default addressing
scheme, to be used only if the URL does not specify one. The default value for
this argument is the empty string.
@ -157,7 +157,7 @@ The :mod:`urlparse` module defines the following functions:
equivalent).
.. function:: urlsplit(urlstring[, default_scheme[, allow_fragments]])
.. function:: urlsplit(urlstring[, scheme[, allow_fragments]])
This is similar to :func:`urlparse`, but does not split the params from the URL.
This should generally be used instead of :func:`urlparse` if the more recent URL