Fixed #21711 -- Enforced unicity of model names.
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@ -185,7 +185,6 @@ class Apps(object):
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# call get_app_config().
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model_name = model._meta.model_name
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app_models = self.all_models[app_label]
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# Defensive check for extra safety.
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if model_name in app_models:
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raise RuntimeError(
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"Conflicting '%s' models in application '%s': %s and %s." %
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@ -162,12 +162,6 @@ class ModelBase(type):
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new_class._default_manager = new_class._default_manager._copy_to_model(new_class)
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new_class._base_manager = new_class._base_manager._copy_to_model(new_class)
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# Bail out early if we have already created this class.
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try:
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return new_class._meta.apps.get_registered_model(new_class._meta.app_label, name)
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except LookupError:
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pass
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# Add all attributes to the class.
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for obj_name, obj in attrs.items():
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new_class.add_to_class(obj_name, obj)
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@ -285,13 +279,8 @@ class ModelBase(type):
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return new_class
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new_class._prepare()
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new_class._meta.apps.register_model(new_class._meta.app_label, new_class)
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# Because of the way imports happen (recursively), we may or may not be
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# the first time this model tries to register with the framework. There
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# should only be one class for each model, so we always return the
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# registered version.
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return new_class._meta.apps.get_registered_model(new_class._meta.app_label, name)
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return new_class
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def copy_managers(cls, base_managers):
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# This is in-place sorting of an Options attribute, but that's fine.
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@ -7,6 +7,7 @@ circular import difficulties.
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"""
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from __future__ import unicode_literals
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from django.apps import apps
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from django.db.backends import utils
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from django.utils import six
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from django.utils import tree
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@ -185,23 +186,29 @@ def deferred_class_factory(model, attrs):
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being replaced with DeferredAttribute objects. The "pk_value" ties the
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deferred attributes to a particular instance of the model.
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"""
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class Meta:
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proxy = True
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app_label = model._meta.app_label
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# The app registry wants a unique name for each model, otherwise the new
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# class won't be created (we get an old one back). Therefore, we generate
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# class won't be created (we get an exception). Therefore, we generate
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# the name using the passed in attrs. It's OK to reuse an existing class
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# object if the attrs are identical.
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name = "%s_Deferred_%s" % (model.__name__, '_'.join(sorted(list(attrs))))
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name = utils.truncate_name(name, 80, 32)
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try:
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return apps.get_model(model._meta.app_label, name)
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except LookupError:
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class Meta:
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proxy = True
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app_label = model._meta.app_label
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overrides = dict((attr, DeferredAttribute(attr, model)) for attr in attrs)
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overrides["Meta"] = Meta
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overrides["__module__"] = model.__module__
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overrides["_deferred"] = True
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return type(str(name), (model,), overrides)
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# The above function is also used to unpickle model instances with deferred
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# fields.
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deferred_class_factory.__safe_for_unpickling__ = True
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@ -664,6 +664,12 @@ If you have two apps with the same label, you should create an
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:class:`~django.apps.AppConfig.label` there. You should then adjust your code
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wherever it references this application or its models with the old label.
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It isn't possible to import the same model twice through different paths any
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more. As of Django 1.6, this may happen only if you're manually putting a
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directory and a subdirectory on :envvar:`PYTHONPATH`. Refer to the section on
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the new project layout in the :doc:`1.4 release notes </releases/1.4>` for
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migration instructions.
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You should make sure that your project doesn't import models from applications
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that aren't in :setting:`INSTALLED_APPS`. Relations involving such models may
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not be created properly. Future versions of Django may forbid this entirely.
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