Decode the non-english username str to unicode

convert the non-english username str to unicode,
then the converted username can match the username from db.

Fixes bug #1221576

Change-Id: I9e5f941fe43f081d75750e3f4754e8beea8210db

keystoneclient/utils.py

@@ -99,6 +99,8 @@ def find_resource(manager, name_or_id):
 
     # finally try to find entity by name
     try:
+        if isinstance(name_or_id, str):
+            name_or_id = name_or_id.decode('utf-8', 'strict')
         return manager.find(name=name_or_id)
     except exceptions.NotFound:
         msg = ("No %s with a name or ID of '%s' exists." %

tests/test_utils.py

@@ -28,6 +28,7 @@ class FakeManager(object):
     resources = {
         '1234': {'name': 'entity_one'},
         '8e8ec658-c7b0-4243-bdf8-6f7f2952c0d0': {'name': 'entity_two'},
+        '\xe3\x82\xbdtest': {'name': u'\u30bdtest'},
         '5678': {'name': '9876'}
     }
 
@@ -72,6 +73,11 @@ class FindResourceTestCase(test_utils.TestCase):
         output = utils.find_resource(self.manager, uuid)
         self.assertEqual(output, self.manager.resources[uuid])
 
+    def test_find_by_unicode(self):
+        name = '\xe3\x82\xbdtest'
+        output = utils.find_resource(self.manager, name)
+        self.assertEqual(output, self.manager.resources[name])
+
     def test_find_by_str_name(self):
         output = utils.find_resource(self.manager, 'entity_one')
         self.assertEqual(output, self.manager.resources['1234'])