Fix up how the JSON Schema compiler decides whether to include or forward

tools/json_schema_compiler/schema_util.py

 # Copyright (c) 2012 The Chromium Authors. All rights reserved.
 # Use of this source code is governed by a BSD-style license that can be
 # found in the LICENSE file.
 """Utilies for the processing of schema python structures.
 """
 
 import json_parse
 
 def CapitalizeFirstLetter(value):
   return value[0].capitalize() + value[1:]
 
-def GetNamespace(ref_type):
-  if '.' in ref_type:
-    return ref_type[:ref_type.rindex('.')]
+def GetNamespace(ref):
+  return SplitNamespace(ref)[0]
 
-def StripSchemaNamespace(s):
-  last_dot = s.rfind('.')
-  if not last_dot == -1:
-    return s[last_dot + 1:]
-  return s
+def StripNamespace(ref):
+  return SplitNamespace(ref)[1]
+
+def SplitNamespace(ref):
+  """Returns (namespace, thing) from |ref|. If |ref| isn't qualified then

[Yoyo Zhou, 2013/01/26 01:43:02]

Surely there's a better word than 'thing'.

[not at google - send to devlin, 2013/01/26 04:35:31]

probably

+  returns (None, ref).
+  """
+  if '.' in ref:
+    return tuple(ref.rsplit('.', 1))
+  return (None, ref)
 
 def JsFunctionNameToClassName(namespace_name, function_name):
   """Transform a fully qualified function name like foo.bar.baz into FooBarBaz
 
   Also strips any leading 'Experimental' prefix."""
   parts = []
   full_name = namespace_name + "." + function_name
   for part in full_name.split("."):
     parts.append(CapitalizeFirstLetter(part))
   if parts[0] == "Experimental":
     del parts[0]
   class_name = "".join(parts)
   return class_name