Changing User ModelAdmin for Django admin

Question

How do you override the admin model for Users? I thought this would work but it doesn't?

class UserAdmin(admin.ModelAdmin):
    list_display = ('email', 'first_name', 'last_name')
    list_filter = ('is_staff', 'is_superuser')

admin.site.register(User, UserAdmin)

I'm not looking to override the template, just change the displayed fields & ordering.

Solutions please?

Answer 1

You have to unregister User first:

class UserAdmin(admin.ModelAdmin):
    list_display = ('email', 'first_name', 'last_name')
    list_filter = ('is_staff', 'is_superuser')


admin.site.unregister(User)
admin.site.register(User, UserAdmin)

Maybe this question is also interesting for you: Customizing an Admin form in Django while also using autodiscover

Answer 2

Update tested with Django 3+

So you don't lose data like password encryption and the form itself, perform the import below

from django.contrib.auth.admin import UserAdmin

Define an AdminCustom class as the example and customize with the options you want, overriding the default.

class UserAdminCustom(UserAdmin):
   list_display = ('email', 'first_name', 'last_name', 'is_staff', 'is_superuser')
   list_filter = ('is_staff', 'is_superuser')
   search_fields = ('username', )

And in the end, follow the example mentioned

admin.site.unregister(User)
admin.site.register(User, UserAdminCustom)

Answer 3

As pointed by @haifeng-zhang in the comments, it is useful to extend the default UserAdmin instead.

The official documentation about this can be found here: https://docs.djangoproject.com/en/dev/topics/auth/customizing/#extending-the-existing-user-model

from django.contrib import admin
from django.contrib.auth.admin import UserAdmin as BaseUserAdmin
from django.contrib.auth.models import User


# Define a new User admin
class UserAdmin(BaseUserAdmin):
    list_display = ('email', 'first_name', 'last_name')
    list_filter = ('is_staff', 'is_superuser')

# Re-register UserAdmin
admin.site.unregister(User)
admin.site.register(User, UserAdmin)