Fix urlunparse for salt file urls

Previously, the `salt.utils.url.create()` method naively assumed the
compiled URL provided by `urllib.parse.urlunparse()` always starts with
the literal string `file:///`, that is, the begininning of the URL
string starting with the `scheme` and including an empty `authority`
section.  Beginning with
[python-3.12](python/cpython@0edfc66),
however, generating a URL with `urlunparse()` results in the syntax
`file:` for the portion of the URL including the `scheme` and
`authority` sections.

Since `urlunparse()` accepts the arbitrary scheme `salt`, use it instead
and then normalize the rendered scheme to `salt://` as SaltStack
expects.

salt/utils/url.py

@@ -46,8 +46,15 @@ def create(path, saltenv=None):
     path = salt.utils.data.decode(path)
 
     query = f"saltenv={saltenv}" if saltenv else ""
-    url = salt.utils.data.decode(urlunparse(("file", "", path, "", query, "")))
-    return "salt://{}".format(url[len("file:///") :])
+    url = salt.utils.data.decode(urlunparse(("salt", "", path, "", query, "")))
+    # urllib may return `salt:` or `salt:///`, but salt wants `salt://`, so normalize here
+    if url.startswith("salt:{path}"):
+        return f"salt://{url.split('salt:')[1]}"
+    elif url.startswith("salt://{path}"):
+        return url
+    elif url.startswith("salt:///{path}"):
+        return f"salt://{url.split('salt:///')[1]}"
+    raise ValueError(f"Cannot interpret {url!r} as a 'salt://' URL")
 
 
 def is_escaped(url):