[PATCH stable] Fix Issue 2111, test-schemes unit test failure

[Michael Glassford]

# HG changeset patch
# User Michael Glassford <glassfordmjg at gmail.com>
# Date 1270731213 14400
# Node ID ab12040bcec69f9f6bdfab0670bcc22a445a897a
# Parent  f97b98db6fd1de6c60c1b60254c9240337dadc01
Fix Issue 2111, test-schemes unit test failure.
Recent Pythons (e.g. 2.6.5 and 3.1) introduce a change that causes
urlparse.urlunparse(urlparse.urlparse('x://')) to return 'x:' instead of 'x://'.
Fix url.hidepassword() and url.removeauth() to handle this case.

diff -r f97b98db6fd1 -r ab12040bcec6 mercurial/url.py
--- a/mercurial/url.py	Mon Apr 05 17:48:48 2010 -0500
+++ b/mercurial/url.py	Thu Apr 08 08:53:33 2010 -0400
@@ -11,17 +11,35 @@
 from i18n import _
 import keepalive, util
 
+def _urlunparse(scheme, netloc, path, params, query, fragment, url):
+    # Recent Pythons (e.g. 2.6.5 and 3.1) introduce a change that 
+    # causes urlparse.urlunparse(urlparse.urlparse('x://')) to 
+    # return 'x:' instead of 'x://' and 
+    # urlparse.urlunparse(urlparse.urlparse('x:///y')) to 
+    # return 'x:/y' instead of 'x:///y',
+    # unless "x" appears in the list urlparse.uses_netloc.
+    # Since we know the original url, re-add the missing "//" 
+    # if the original url had it.
+    result = urlparse.urlunparse((scheme, netloc, path, params, query, fragment))
+    if (scheme and
+        result.startswith(scheme + ':') and
+        not result.startswith(scheme + '://') and
+        url.startswith(scheme + '://')
+       ):
+        result = scheme + '://' + result[len(scheme + ':'):]
+    return result
+
 def hidepassword(url):
     '''hide user credential in a url string'''
     scheme, netloc, path, params, query, fragment = urlparse.urlparse(url)
     netloc = re.sub('([^:]*):([^@]*)@(.*)', r'\1:***@\3', netloc)
-    return urlparse.urlunparse((scheme, netloc, path, params, query, fragment))
+    return _urlunparse(scheme, netloc, path, params, query, fragment, url)
 
 def removeauth(url):
     '''remove all authentication information from a url string'''
     scheme, netloc, path, params, query, fragment = urlparse.urlparse(url)
     netloc = netloc[netloc.find('@')+1:]
-    return urlparse.urlunparse((scheme, netloc, path, params, query, fragment))
+    return _urlunparse(scheme, netloc, path, params, query, fragment, url)
 
 def netlocsplit(netloc):
     '''split [user[:passwd]@]host[:port] into 4-tuple.'''