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如何在Django基于类的视图中使用permission_required装饰器

djangodjango-viewsdjango-authenticationdjango-class-based-views
186
我有一些难以理解新的类视图（Class Based Views）是如何工作的。我的问题是，我需要在所有视图中要求登录，在某些视图中还需要特定权限。在基于函数的视图中，我使用@permission_required()和视图中的login_required属性来完成这个操作，但是我不知道如何在新的视图上实现这个功能。Django文档中有关于这方面的介绍吗？我没有找到任何相关内容。我的代码哪里出了问题呢？
我尝试使用@method_decorator，但它回复错误：TypeError at /spaces/prueba/ _wrapped_view() takes at least 1 argument (0 given) 以下是代码（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')
        context['publication'] = Post.objects.filter(post_space=place.id).order_by('-post_pubdate')
        return context
13个回答
1
我根据Josh的解决方案进行了修复。
class LoginRequiredMixin(object):

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

使用示例：

class EventsListView(LoginRequiredMixin, ListView):

    template_name = "events/list_events.html"
    model = Event
0
如果您正在进行需要各种权限测试的项目，您可以继承此类。
from django.contrib.auth.decorators import login_required
from django.contrib.auth.decorators import user_passes_test
from django.views.generic import View
from django.utils.decorators import method_decorator



class UserPassesTest(View):

    '''
    Abstract base class for all views which require permission check.
    '''


    requires_login = True
    requires_superuser = False
    login_url = '/login/'

    permission_checker = None
    # Pass your custom decorator to the 'permission_checker'
    # If you have a custom permission test


    @method_decorator(self.get_permission())
    def dispatch(self, *args, **kwargs):
        return super(UserPassesTest, self).dispatch(*args, **kwargs)


    def get_permission(self):

        '''
        Returns the decorator for permission check
        '''

        if self.permission_checker:
            return self.permission_checker

        if requires_superuser and not self.requires_login:
            raise RuntimeError((
                'You have assigned requires_login as False'
                'and requires_superuser as True.'
                "  Don't do that!"
            ))

        elif requires_login and not requires_superuser:
            return login_required(login_url=self.login_url)

        elif requires_superuser:
            return user_passes_test(lambda u:u.is_superuser,
                                    login_url=self.login_url)

        else:
            return user_passes_test(lambda u:True)
0

这里是permission_required修饰器的解决方案：

class CustomerDetailView(generics.GenericAPIView):

@method_decorator(permission_required('app_name.permission_codename', raise_exception=True))
    def post(self, request):
        # code...
        return True
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