Commit fd702fe0 authored by Malcolm Tredinnick's avatar Malcolm Tredinnick
Browse files

Fixed #1796 -- only load a single copy of each model, even when it is 

referenced via different import paths. Solves a problem with many-to-many
relations being set up incorrectly. Thanks to Curtis Thompson, Luke Plant and
Simon Willison for some excellent debugging of the problem. Refs #2232 (may
also have fixed that).


git-svn-id: http://code.djangoproject.com/svn/django/trunk@3202 bcc190cf-cafb-0310-a4f2-bffc1f526a37
parent 23c24fc0

django/db/models/loading.py

+12 −0
Original line number Diff line number Diff line
@@ -2,6 +2,8 @@ 


from django.conf import settings

from django.core.exceptions import ImproperlyConfigured

import sys

import os



__all__ = ('get_apps', 'get_app', 'get_models', 'get_model', 'register_models')
@@ -91,4 +93,14 @@ def register_models(app_label, *models): 
        # in the _app_models dictionary

        model_name = model._meta.object_name.lower()

        model_dict = _app_models.setdefault(app_label, {})

        if model_dict.has_key(model_name):

            # The same model may be imported via different paths (e.g.

            # appname.models and project.appname.models). We use the source

            # filename as a means to detect identity.

            fname1 = os.path.normpath(sys.modules[model.__module__].__file__)

            fname2 = os.path.normpath(sys.modules[model_dict[model_name].__module__].__file__)

            # Since the filename extension could be .py the first time and .pyc

            # or .pyo the second time, ignore the extension when comparing.

            if os.path.splitext(fname1)[0] == os.path.splitext(fname2)[0]:

                continue

        model_dict[model_name] = model