Django: how to route URL requests to django.contrib.admin functions - django

Question

I would like Django to return the same response for requests to /myapp/add as /admin/myapp/mymodel/add.
myproject/myapp/models.py defines the model and myapp/admin.py registers with django.contrib.admin.
myproject/myapp/models.py:
from django.db import models
class MyModel(models.Model):
...
myproject/myapp/admin.py:
from django.contrib import admin
from .models import MyModel
admin.site.register(MyModel)
I am stuck on how to route the request to django.contrib.admin in the project's urlpatterns:
myproject/myproject/urls.py:
urlpatterns = [
url(r'^$', views.home_page, name='home'),
url(r'^admin/', admin.site.urls),
url(r'^myapp/add', ??????),
]
From printing the return from resolve('/admin/myapp/mymodel/add/') this looks like part of the answer:
ResolverMatch(func=django.contrib.admin.options.add_view, args=(), kwargs={}, url_name=myapp_mymodel_add, app_names=['admin'], namespaces=['admin'])

Answer 1

Well let me say that seems like a weird thing to do, but anyway:
in the file: django.contrib.admin.options.py we see:
class ModelAdmin(BaseModelAdmin):
we see that the add view returns
_changeform_view()
which uses the template
django/contrib/admin/templates/change_form.html
So you would want to render that template in your view.
but it would be missing a bunch of context items,
so you would basically have to re-implement the django.admin.options._changeform_view
and then copy the template django/contrib/admin/templates/change_form.html to your apps' template directory
def admin_add(request):
# admin.changeform_view code here
return render(request, "myapp/change_form.html", {context{)
ps. I assume the admin site view assumes the user is the "superuser" and not a normal user so you would want to account for that as well..