Override method of a django builtin Class - django

Question

Abstract of Question:
I am trying to extend (not replace) some method of any of django's builtin Class in my custom module which would allow to further extend the extended(overridden) method.
Lets say I have two custom modules mod1 and mod2. Want to override same method of some django's class say get_apps_list of AdminSite, In mod1 I want to add a line to say hello, in mod2 it should say hi.
Desired output:
There should be nothing if none of the modules installed,
it should say hi if mod1 installed and hello if mod2
installed.
And hi and hello if both installed
Question with real example:
Just for example, I need to modify the implementation of AdminSite.get_app_list like following
From
#app['models'].sort(key=lambda x: x['name'])
To
app['models'] = sort_with_name_length(app['models']) //my own method
Expected/desired approach: I supposed that it should be achievable by just writing following code in any of my custom module's models.py or sites.py file
class MyAdminSite(AdminSite):
def get_app_list(self, request):
app_list = super().get_app_list(request)
for app in app_list:
#app['models'].sort(key=lambda x: x['name'])
app['models'] = sort_with_name_length(app['models']) #an example change I need
return app_list
But what above code does is nothing, its never executed, until I use Monkey patching guided by this answer.
What I could achieve
from django.contrib.admin import AdminSite
class MyAdminSite(AdminSite):
def get_app_list(self, request):
# res = super(MyAdminSite, self).get_app_list(request) //gives following error
# super(type, obj): obj must be an instance or subtype of type
# So i have to rewrite complete method again in my module like following
app_dict = self._build_app_dict(request)
app_list = sorted(app_dict.values(), key=lambda x: x['name'].lower())
for app in app_list:
#app['models'].sort(key=lambda x: x['name'])
app['models'] = sort_with_name_length(app['models'])
return app_list
AdminSite.get_app_list = MyAdminSite.get_app_list
Problem being faced: Above does what I need in a totally undesired way. This solution has two problems
It will not allow me Multilevel Inheritance (I would not be able to have child and grand child)
Its actually even not an overriding, its just a replacement of implementation as it gives error using super
Just for elaboration, following is an example of similar behavior overriding in odoo, what I want to achieve with django
The exact expected/desired behavior is offered by odoo.
You can see get_auth_signup_qcontext method of auth_oauth's main controller
https://github.com/odoo/odoo/blob/14.0/addons/auth_oauth/controllers/main.py
https://github.com/odoo/odoo/blob/14.0/addons/auth_signup/controllers/main.py
What is does is if auth_oauth module is installed anywhere we call get_auth_signup_qcontext it would first go to child(auth_oauth)'s get_auth_signup_qcontext method which will call super in it. But if auth_oauth is not installed anywhere we call get_auth_signup_qcontext will directly hit auth_signup's method

Answer 1

Method 1. Use your own admin site's urls
One can simply use their own admin site's urls (reference):
from django.contrib.admin import AdminSite
class MyAdminSite(AdminSite):
def get_app_list(self, request):
app_list = super().get_app_list(request)
for app in app_list:
#app['models'].sort(key=lambda x: x['name'])
app['models'] = sort_with_name_length(app['models']) #an example change I need
return app_list
admin_site = MyAdminSite(name='myadmin')
admin_site.register(MyModel) # registering models
Then in your urls:
from django.urls import path
from myapp.admin import admin_site
urlpatterns = [
path('myadmin/', admin_site.urls),
]
Method 2. Overriding the default admin site
You can override the admin site:
In apps.py:
from django.contrib.admin.apps import AdminConfig
class MyAdminConfig(AdminConfig):
default_site = 'myproject.admin.MyAdminSite'
In settings.py:
INSTALLED_APPS = [
...
'myproject.apps.MyAdminConfig', # replaces 'django.contrib.admin'
...
]

Answer 2

Quite an awkward and hacky but completely independent/loosely coupled solution to really override/extend (not replace) a builtin class of django's method is
Caution read the complete answer before using code :)
Add following code to models.py of any of your custom (installed) app
from django.contrib.admin import AdminSite
original_get_app_list = AdminSite.get_app_list
class AdminSiteExtension1(AdminSite):
def get_app_list(self, request):
// do something here to manipulate earlier than calling parent
// Following is the solution line
app_list = original_get_app_list(self, request)
for app in app_list:
app['name'] = app['name'] + '_my_app'
return app_list
AdminSite.get_app_list = AdminSiteExtension1.get_app_list
AdminSite is inherited just to make the use of self
Its really loosely coupled as u need to do nothing anywhere else. Even you can extend it further in any other module and you will not need AdminSiteExtension1 in that module because the updates to get_app_list of original AdminSite reside in original method as long as the app having AdminSiteExtension1 is installed.
Disclosure: I am not really a well learned and visionary programmer so I cannot imagine a situation where this solution can cause any problem, so if someone guides, it would be welcome. otherwise the beneficiary has to take care him/herself.