gh-129745: Fix urlparse example to properly parse params

Changed the example from using 'scheme://' to 'http://' so that the
params component is correctly parsed. The generic 'scheme' isn't in
the list of schemes that support params, so the example was showing
params remaining in the path rather than being extracted.

Co-Authored-By: Claude Opus 4.5 <noreply@anthropic.com>

Author: kovan

Doc/library/urllib.parse.rst

@@ -69,8 +69,8 @@ or on combining URL components into a URL string.
       :options: +NORMALIZE_WHITESPACE
 
       >>> from urllib.parse import urlparse
-      >>> urlparse("scheme://netloc/path;parameters?query#fragment")
-      ParseResult(scheme='scheme', netloc='netloc', path='/path;parameters', params='',
+      >>> urlparse("http://netloc/path;parameters?query#fragment")
+      ParseResult(scheme='http', netloc='netloc', path='/path', params='parameters',
                   query='query', fragment='fragment')
       >>> o = urlparse("http://docs.python.org:80/3/library/urllib.parse.html?"
       ...              "highlight=params#url-parsing")