attempt to find resource by listing

Steve Martinelli · Dean Troyer · commit fe2d9f479fc8 · 2015-09-30T13:19:56.000-05:00
add a last-ditch effort to find the resource in question by listing all the resources and doing a simply match for name and id. if no match is found then raise an error, if the list call is unsuccessful, raise the same error. we have failed this city. Closes-Bug: #1501362 Change-Id: I0d3d7002e9ac47b17b1ef1a5534406c85b1fc753 (cherry picked from commit 83282bc)
diff --git a/openstackclient/common/utils.py b/openstackclient/common/utils.py
@@ -121,8 +121,24 @@ def find_resource(manager, name_or_id, **kwargs):
                 (manager.resource_class.__name__.lower(), name_or_id)
             raise exceptions.CommandError(msg)
         else:
-            msg = "Could not find resource %s" % name_or_id
-            raise exceptions.CommandError(msg)
+            pass
+
+    try:
+        for resource in manager.list():
+            # short circuit and return the first match
+            if (resource.get('id') == name_or_id or
+                    resource.get('name') == name_or_id):
+                return resource
+        else:
+            # we found no match, keep going to bomb out
+            pass
+    except Exception:
+        # in case the list fails for some reason
+        pass
+
+    # if we hit here, we've failed, report back this error:
+    msg = "Could not find resource %s" % name_or_id
+    raise exceptions.CommandError(msg)
 
 
 def format_dict(data):
