Django 1.9 deprecation warnings app_label

Add a meta class to your model with an app_label attribute.

class Meta:
    app_label = 'app_model_belongs_to'

Hope this works!

EDIT:
The reason for this is usually that the model exists outside of the standard locations.

For more information refer to:
https://docs.djangoproject.com/en/1.8/ref/models/options/#app-label

I had the same problem. I put a breakpoint in django.db.models.base.py:line82 and try to figure out what is causing this warning message.

# Look for an application configuration to attach the model to.
app_config = apps.get_containing_app_config(module)

Basically, if your app does not exist by this time, you get that warning. I realized that my problem was that I have a third party framework (in my case, haystack) that tries to import one of my custom models.

Maybe you also have a third party package listed in INSTALLED_APPS before your custom apps and your third party package references your custom apps? This would be possible if you are using something like Django rest framework as well.

I was getting a similar error, but rather than complaining about models in my apps, it was complaining about models in contrib packages. For example:

C:Program FilesPython 2.7libsite-packagesdjangocontribsessionsmodels.py:27: RemovedInDjango19Warning: Model class django.contrib.sessions.models.Session doesn’t declare an explicit app_label and either isn’t in an application in INSTALLED_APPS or else was imported before its application was loaded. This will no longer be supported in Django 1.9.
class Session(models.Model):

This is caused by a bad ordering in the INSTALLED_APPS property in settings.py. My settings.py initially contained:

INSTALLED_APPS = (
    'my_app_1',
    'my_app_2',
    'my_app_3',
    'bootstrap_admin',
    'django.contrib.admin',
    'django.contrib.auth',
    'django.contrib.contenttypes',
    'django.contrib.messages',
    'django.contrib.sessions',
    'django.contrib.sites',
    'django.contrib.staticfiles',
    'social.apps.django_app.default',
    'mathfilters',
    'rest_framework',
)

my_app_* use models from the contrib packages. The error is caused by using models before declaring them (i.e. Django should know about the apps containing those models before using them).

In order to solve this, the order in which the apps are declared needs to be changed. Specifically, all Django apps should come before the user defined apps. In my case, the correct INSTALLED_APPS would look like:

INSTALLED_APPS = (
    'bootstrap_admin',
    'django.contrib.admin',
    'django.contrib.auth',
    'django.contrib.contenttypes',
    'django.contrib.messages',
    'django.contrib.sessions',
    'django.contrib.sites',
    'django.contrib.staticfiles',
    'social.apps.django_app.default',
    'mathfilters',
    'rest_framework',
    'my_app_1',
    'my_app_2',
    'my_app_3',
)

Now I know this might not directly answer your question, but it answers a related one, and since this is the only SO link that shows on Google when pasting the error, I have answered it here.

However, I believe a similar situation is causing your problem:

Make sure you declare the “dependency” apps before the apps using them! The errors don’t really specify which app is using a model, so you would have to push the app containing the model it mentions to the top, one by one, until the error disappears.

Similar error. In my case the error was:

RemovedInDjango19Warning: Model class django.contrib.sites.models.Site doesn't declare an explicit app_label and either isn't in an application in INSTALLED_APPS or else was imported before its application was loaded. This will no longer be supported in Django 1.9.
class Site(models.Model):

My solution was:

Added 'django.contrib.sites' to INSTALLED_APPS

The Django 1.9 way of handling this and giving your app a nice name in the admin is to do the following:

Add a file in your app named apps.py and add the following to it:

#apps.py
from django.apps import AppConfig


class YourAppNameAppConfig(AppConfig):
    name = 'yourappname'
    verbose_name = 'Your App Name Looking Right'

Then, in your app’s __init__.py file, add the following:

#__init__.py    
default_app_config = 'youappname.apps.YourAppNameAppConfig'

I faced this issue as well and it was related to the way I was loading the signals.py module from an app.

As you may now, it’s quite common to have circular import issues between the signals and the models, and it’s usual to import them from the app’s __init__.py file or from the bottom of the models.py file to avoid them.

Well, I was using the __init__.py approach, and just moving the import signals statement to the bottom of my models.py file solved the issue.

Hope this helps someone else!

I has this problem after upgrade Django from 1.8 to 1.9.1:

RuntimeError at /

Model class blog.models.BlogCategory doesn’t declare an explicit app_label and isn’t in an application in INSTALLED_APPS.

This help to resolve:

in blog/models.py:

class BlogCategory(models.Model):
    some vars & methods

    class Meta:
        app_label = 'BlogCategory'

It’s work on 100%.

I didn’t receive any error in django 1.7.
While migrating to django 1.8 I got this error.
Reason was the signal were defined in app/signal_receiver.py.

I created a apps.py

from django.apps import AppConfig

class TasksConfig(AppConfig):
    name = 'core'
    verbose_name = "core"

    def ready(self):
        import core.signal.handler

I moved signal receiver in handler.py inside signal package.

The easiest and simplest way to solve this is to place the “import signals” command at the BOTTOM of the Model file you are working with. This ensures that the models are all loaded before the signals for that model are imported. You would have to do this for each model you are importing (if you use receivers linked to particular models), or in the models.py for the app that comes at the end of the “installed apps” in your settings.

The other solutions are only required if you are dealing with non-model type signals, where the models will never be imported first.

Why this is not noted in the docs is a mystery, because I just got caught by it, until I remembered that you have to do that.

models.py

from django.db import models

class MyModel(models.Model):
    myfield1 = models.CharField()
    myfield2 = models.CharField()

import signals

And then in signals.py:

from django.db.models.signals import pre_save # Or whatever you are using
from django.dispatch import receiver

from .models import MyModel

@receiver(pre_save, sender=MyModel)
def my_receiver(sender, instance, **kwargs):
    mysender = sender
    print mysender

Like that.

I suspect it’ll be only a tiny minority of people who see this error for whom it will be caused by the same thing as me, but in case it helps someone else it seems worth adding this answer!

I suddenly saw lots of these errors when running tests at one point – it turned out that I had accidentally created a __init__.py at the top level of my Django project when it was meant to be in a subdirectory. The clue that this was happening is that the errors, which were like:

/home/mark/mystupiddjangoproject/alerts/models.py:15: RemovedInDjango19Warning: Model class mystupiddjangoproject.alerts.models.Alert doesn't declare an explicit app_label and either isn't in an application in INSTALLED_APPS or else was imported before its application was loaded. This will no longer be supported in Django 1.9.
  class Alert(models.Model):

… included the name of directory the project is in (mystupiddjangoproject) in the fully qualified model name, which should have been: alerts.models.Alert.

To fix this, I just needed to do:

rm __init__.py
rm __init__.pyc

Sometimes an invalid .pyc files that doesnt match with the current source code caused this problem.

I removed all .pyc to refresh all of them using this bash


find . -name "*.pyc" -exec rm -rf {} ;


I got this error because do with error:

urlpatterns = patterns(
url(r'^my_link/$', views.MyView.as_view(), name='my_view'),...

Do right:

urlpatterns = patterns(
'', # Add this line
url(r'^my_link/$', views.MyView.as_view(), name='my_view'),
...)