超级用户所需的装饰符

小开

最佳答案

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):
...

小开

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):
...

小开

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.

小开

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):
# ...

小开

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):

小开

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):
#...

小开

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):
...
...
...

小开

To check if user is logged in use @login_required decorator and check if logged in user is superuser or not inside the function through if/else condition and return your response accordingly.

'''

    from django.shortcuts import HttpResponse, redirect
from django.contrib.auth.decorators import login_required




@login_required
def function_name(request):
if not request.user.is_superuser:
return redirect('profile')
else:
return HttpResponse('Superuser')

'''