Alex Rivera | Logout

How to use permission_required decorators on django class-based views

Asked 2011-05-20T08:01:11.850
194

I'm having a bit of trouble understanding how the new CBVs work. My question is this, I need to require login in all the views, and in some of them, specific permissions. In function-based views I do that with @permission_required() and the login_required attribute in the view, but I don't know how to do this on the new views. Is there some section in the django docs explaining this? I didn't found anything. What is wrong in my code?

I tried to use the @method_decorator but it replies "TypeError at /spaces/prueba/ _wrapped_view() takes at least 1 argument (0 given)"

Here is the code (GPL):

from django.utils.decorators import method_decorator
from django.contrib.auth.decorators import login_required, permission_required

class ViewSpaceIndex(DetailView):

    """
    Show the index page of a space. Get various extra contexts to get the
    information for that space.

    The get_object method searches in the user 'spaces' field if the current
    space is allowed, if not, he is redirected to a 'nor allowed' page. 
    """
    context_object_name = 'get_place'
    template_name = 'spaces/space_index.html'

    @method_decorator(login_required)
    def get_object(self):
        space_name = self.kwargs['space_name']

        for i in self.request.user.profile.spaces.all():
            if i.url == space_name:
                return get_object_or_404(Space, url = space_name)

        self.template_name = 'not_allowed.html'
        return get_object_or_404(Space, url = space_name)

    # Get extra context data
    def get_context_data(self, **kwargs):
        context = super(ViewSpaceIndex, self).get_context_data(**kwargs)
        place = get_object_or_404(Space, url=self.kwargs['space_name'])
        context['entities'] = Entity.objects.filter(space=place.id)
        context['documents'] = Document.objects.filter(space=place.id)
        context['proposals'] = Proposal.objects.filter(space=place.id).order_by('-pub_date'
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1 Answer

120

Here is my approach, I create a mixin that is protected (this is kept in my mixin library):

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

class LoginRequiredMixin(object):
    @method_decorator(login_required)
    def dispatch(self, request, *args, **kwargs):
        return super(LoginRequiredMixin, self).dispatch(request, *args, **kwargs)

Whenever you want a view to be protected you just add the appropriate mixin:

class SomeProtectedViewView(LoginRequiredMixin, TemplateView):
    template_name = 'index.html'

Just make sure that your mixin is first.

Update: I posted this in way back in 2011, starting with version 1.9 Django now includes this and other useful mixins (AccessMixin, PermissionRequiredMixin, UserPassesTestMixin) as standard!

answered 2011-06-23T13:51:52.830

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