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):

I assumescheme isn't a scheme thatuses_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 aboutuses_params. Is that an oversight? Should some mention be added?

Linked PRs

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