Register every table/class from an app in the Django admin page

Question

I have an app named doors and my models.py for the app has 10 tables/class. Under my admin.py, how do I register every model in the file models.py?

For example, currently I have to hardcode it:

from django.contrib import admin
from doors.models import *

admin.site.register(Group)
admin.site.register(Item)
admin.site.register(ItemType)
admin.site.register(Location)
admin.site.register(Log)
admin.site.register(Order)
admin.site.register(Property)
admin.site.register(User)
admin.site.register(Vendor)

Is there a way I perhaps find every class in models.py and loop through and register each class? Or is there some kind of wildcard I can use with Django?

Answer 1

Seems get_models and get_app are no longer available in django 1.8.

The following can be used:

from django.contrib import admin from django.apps import apps  app = apps.get_app_config('dashboard')  for model_name, model in app.models.items():     admin.site.register(model) 

EXTENSION: If you would like to show all or select fields of the model as a grid instead of a single column unicode representation of the model objects you may use this:

app = apps.get_app_config('your_app_name') for model_name, model in app.models.items():     model_admin = type(model_name + "Admin", (admin.ModelAdmin,), {})      model_admin.list_display = model.admin_list_display if hasattr(model, 'admin_list_display') else tuple([field.name for field in model._meta.fields])     model_admin.list_filter = model.admin_list_filter if hasattr(model, 'admin_list_filter') else model_admin.list_display     model_admin.list_display_links = model.admin_list_display_links if hasattr(model, 'admin_list_display_links') else ()     model_admin.list_editable = model.admin_list_editable if hasattr(model, 'admin_list_editable') else ()     model_admin.search_fields = model.admin_search_fields if hasattr(model, 'admin_search_fields') else ()      admin.site.register(model, model_admin) 

What this does is, it extends ModelAdmin class on the fly and sets the list_display field which is required for showing model data in grid representation in the admin. If you list your desired fields in your model as admin_list_display it takes that one, or generates a tuple of all fields available in the model, otherwise.

Other optional fields can similarly be set, such as list_filter.

See django documentation for more info on list_display.

Answer 2

I figured it out with @arie's link (for django < 1.8):

from django.contrib import admin
from django.db.models import get_models, get_app

for model in get_models(get_app('doors')):
    admin.site.register(model)

But I wonder if I can do this without get_app... Couldn't the code be smart enough to know the name of its own app?