django @login_required decorator for a superuser

DjangoDjango Views

Django Problem Overview


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

Thanks

Django Solutions


Solution 1 - Django

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):
    ...

Solution 2 - Django

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):
    ...

Solution 3 - Django

If you want to have similar functionality to @staff_member_required you can easily write your own decorator. Taking @staff_member as an example we can do something like this:

from django.contrib.auth import REDIRECT_FIELD_NAME
from django.contrib.admin.views.decorators import user_passes_test

def superuser_required(view_func=None, redirect_field_name=REDIRECT_FIELD_NAME,
                   login_url='account_login_url'):
    """
    Decorator for views that checks that the user is logged in and is a
    superuser, redirecting to the login page if necessary.
    """
    actual_decorator = user_passes_test(
        lambda u: u.is_active and u.is_superuser,
        login_url=login_url,
        redirect_field_name=redirect_field_name
    )
    if view_func:
        return actual_decorator(view_func)
    return actual_decorator

This example is a modified staff_member_required, just changed one check in lambda.

Solution 4 - Django

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):
        # ...

Solution 5 - Django

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):

Solution 6 - Django

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):
    #...

Solution 7 - Django

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):
    ...
    ...
    ...

Attributions

All content for this solution is sourced from the original question on Stackoverflow.

The content on this page is licensed under the Attribution-ShareAlike 4.0 International (CC BY-SA 4.0) license.

Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestionMadMaardiganView Question on Stackoverflow
Solution 1 - DjangodgelView Answer on Stackoverflow
Solution 2 - DjangoBit68View Answer on Stackoverflow
Solution 3 - DjangokoradonView Answer on Stackoverflow
Solution 4 - DjangoabidiboView Answer on Stackoverflow
Solution 5 - DjangoKamil MarczakView Answer on Stackoverflow
Solution 6 - Djangokn3lView Answer on Stackoverflow
Solution 7 - DjangoRobView Answer on Stackoverflow