22👍
Interesting question, which provoked me to do a little digging.
Once the admin classes have been registered, they are stored in an attribute of the site
object called – not surprisingly – _registry
. This is a dictionary of model classes to modeladmin classes – note both the keys and values are classes, not names.
So if you have an admin.py like this:
from django.contrib import admin
from myapp.models import MyModel
class MyModelAdmin(admin.ModelAdmin):
list_display = ('field1', 'field2')
admin.site.register(MyModel, MyModelAdmin)
then once that has actually been imported – usually by the admin.autodiscover()
line in urls.py – admin.site._registry
will contain something like this:
{<class 'myapp.models.MyModel'>:
<django.contrib.admin.options.ModelAdmin object at 0x10210ba50>}
and you would get the ModelAdmin object for MyModel
by using the model itself as the key:
>>> admin.site._registry[MyModel]
<django.contrib.admin.options.ModelAdmin object at 0x10210ba50>
3👍
Django’s django.contrib.admin.sites.AdminSite
has a method for checking registered Model called .is_registered(model)
. This method will check on the admin site’s _registry
attribute (just like Daniel Roseman‘s approach)
So, if you have files like these:
# models.py
from django.db import models
class MyModel(models.Model)
field1 = ...
field2 = ...
# admin.py
from django.contrib import admin
from .models import MyModel
class MyModelAdmin(admin.ModelAdmin):
list_display = ('field1', 'field2')
admin.site.register(MyModel, MyModelAdmin)
You could make some test like this:
# tests.py
from django.test import TestCase
from .models import MyModel
class TestModelAdmin(TestCase):
def test_mymodel_registered(self):
self.assertTrue(admin.site.is_registered(MyModel))
nb: I’ve checked it on the Django’s modules documentation from Django 1.8 to Django 2.2