The Python Oracle

Multiple ModelAdmins/views for same model in Django admin

--------------------------------------------------
Hire the world's top talent on demand or became one of them at Toptal: https://topt.al/25cXVn
--------------------------------------------------

Music by Eric Matyas
https://www.soundimage.org
Track title: Fantascape Looping

--

Chapters
00:00 Multiple Modeladmins/Views For Same Model In Django Admin
01:50 Accepted Answer Score 364
02:43 Answer 2 Score 3
03:01 Answer 3 Score 2
03:30 Thank you

--

Full question
https://stackoverflow.com/questions/2223...

--

Content licensed under CC BY-SA
https://meta.stackexchange.com/help/lice...

--

Tags
#python #django #djangoadmin

#avk47



ACCEPTED ANSWER

Score 364


I've found one way to achieve what I want, by using proxy models to get around the fact that each model may be registered only once.

class PostAdmin(admin.ModelAdmin):
    list_display = ('title', 'pubdate','user')

class MyPost(Post):
    class Meta:
        proxy = True

class MyPostAdmin(PostAdmin):
    def get_queryset(self, request):
        return self.model.objects.filter(user = request.user)


admin.site.register(Post, PostAdmin)
admin.site.register(MyPost, MyPostAdmin)

Then the default PostAdmin would be accessible at /admin/myapp/post and the list of posts owned by the user would be at /admin/myapp/myposts.

After looking at http://code.djangoproject.com/wiki/DynamicModels, I've come up with the following function utility function to do the same thing:

def create_modeladmin(modeladmin, model, name = None):
    class  Meta:
        proxy = True
        app_label = model._meta.app_label

    attrs = {'__module__': '', 'Meta': Meta}

    newmodel = type(name, (model,), attrs)

    admin.site.register(newmodel, modeladmin)
    return modeladmin

This can be used as follows:

class MyPostAdmin(PostAdmin):
    def get_queryset(self, request):
        return self.model.objects.filter(user = request.user)

create_modeladmin(MyPostAdmin, name='my-posts', model=Post)



ANSWER 2

Score 3


Paul Stone answer is absolutely great! Just to add, for Django 1.4.5 I needed to inherit my custom class from admin.ModelAdmin

class MyPostAdmin(admin.ModelAdmin):
    def queryset(self, request):
        return self.model.objects.filter(id=1)



ANSWER 3

Score 2


Based on the correct answers, I monkeypatch the AdminSite class and add the method register_via_proxy to make the task easier.

import re
from django.contrib import admin

def _register_proxy(self, model, admin_class):
    proxy_model = type(
        admin_class.__name__, (model,), {
            "__module__": re.sub(
                r'(^.*?)(\.[^\.]+)$', r'\1.proxy', model.__module__
            ),
            "Meta": type("Meta", tuple(), {
                "proxy": True,
                 "app_label": model._meta.app_label
            })
        }
    )

    return self.register(proxy_model, admin_class)


admin.sites.AdminSite.register_via_proxy = _register_proxy

And to use is like:

site = admin.sites.AdminSite()
site.register_via_proxy(models.ModelType, AdminClass)