django @login_required decorator for a superuser

Question

Is there a decorator in django similar to @login_required that also tests if the user is a superuser?

Thanks

Answer 1

In case staff membership is sufficient and you do not need to check whether the user is a superuser, you can use the @staff_member_required decorator:

from django.contrib.admin.views.decorators import staff_member_required

@staff_member_required
def my_view(request):
    ...

Answer 2

if you have your profile of user you can simply do this

@login_required
@user_passes_test(lambda u: True if u.profile.role==2 else False )
def add_listing(request):
    #...

Answer 3

To require a superuser on a class based view without writing new code:

from django.utils.decorators import method_decorator
from django.contrib.auth.decorators import user_passes_test

@method_decorator(user_passes_test(lambda u: u.is_superuser), name='dispatch')
class AdminCreateUserView(LoginRequiredMixin, FormView):
    ...
    ...
    ...

Answer 4

For class based views, creating a reusable decorator:

from django.contrib.auth.mixins import UserPassesTestMixin
from django.views.generic import View


def superuser_required():
    def wrapper(wrapped):
        class WrappedClass(UserPassesTestMixin, wrapped):
            def test_func(self):
                return self.request.user.is_superuser

        return WrappedClass
    return wrapper

@superuser_required()
class MyClassBasedView(View):
    def get(self, request):
        # ...

Answer 5

Use the user_passes_test decorator:

from django.contrib.auth.decorators import user_passes_test

@user_passes_test(lambda u: u.is_superuser)
def my_view(request):
    ...

Answer 6

I recommend using Mixins, example:

from django.contrib.auth.mixins import UserPassesTestMixin


class SuperUserCheck(UserPassesTestMixin, View):
    def test_func(self):
        return self.request.user.is_superuser

Then you can add SuperUserCheck to View class:

class MyView(SuperUserCheck, View):