DOC: `urllib.parse.urlparse()` example doesn't parse the `params`

[wjandrea]

https://docs.python.org/3/library/urllib.parse.html#urllib.parse.urlparse

>>> urlparse("scheme://netloc/path;parameters?query#fragment")
ParseResult(scheme='scheme', netloc='netloc', path='/path;parameters', params='',
            query='query', fragment='fragment')

I think the problem is here (for v3.13):

cpython/Lib/urllib/parse.py

Line 401 in 78377c7

if (scheme or '') in uses_params and ';' in url:

I assume scheme isn't a scheme that uses_params, so the params don't get parsed out. On the other hand, https works as expected:

>>> urlparse("https://netloc/path;parameters?query#fragment")
ParseResult(scheme='https', netloc='netloc', path='/path', params='parameters',
            query='query', fragment='fragment')

By the way, the documentation doesn't mention anything about uses_params. Is that an oversight? Should some mention be added?

Linked PRs

• gh-129745: Fix urlparse example to properly parse params #144461

[vadmium]

A list of supported schemes is given in the second paragraph of the page. A scheme literally named “scheme:” is not listed, so is confusing as an example.

I think uses_params is just an implementation detail (of the documented scheme list) that unfortunately has a public-looking name.

[wjandrea]

I think uses_params is just an implementation detail (of the documented scheme list)

No, uses_params is smaller than the documented scheme list, e.g. gopher is missing.

[ncoghlan]

Also see #144148 (urlparse has assorted legacy quirks that urlsplit fixes, so the latter API should generally be preferred in new code, but the docs don't present the two functions that way)

[serhiy-storchaka]

This example has been removed.