Django REST框架GenericAPIView权限
Django REST框架GenericAPIView权限
提问于 2018-06-11 04:55:17
我试图在django rest框架中为我的视图添加权限，在类中为特定的视图添加权限，permissions_classes就像这样：
from rest_framework.permissions import IsAuthenticated
from rest_framework.response import Response
from rest_framework.views import APIView
class ExampleView(APIView):
permission_classes = (IsAuthenticated,)
def get(self, request, format=None):
content = {
'status': 'request was permitted'
}
return Response(content)
我有一个略有不同的观点，我generics.GenericAPIView试图用这种方式设置权限时使用并且遇到了一个障碍：
from games.models import Game
from games.serializers import GameSerializer
from rest_framework import mixins
from rest_framework import generics
from rest_framework.permissions import IsAuthenticated
class GameList(mixins.ListModelMixin, generics.GenericAPIView):
permission_classes = (IsAuthenticated,)
queryset = Game.objects.all()
serializer_class = GameSerializer
def get(self, request, *args, **kwargs):
return self.list(request, *args, **kwargs)
class GameDetail(mixins.RetrieveModelMixin, generics.GenericAPIView):
queryset = Game.objects.all()
serializer_class = GameSerializer
def get(self, request, *args, **kwargs):
return self.retrieve(request, *args, **kwargs)
以这种方式设置权限时，API只是简单地停止响应。我在这里设置我的权限不正确吗？
回答 1
Stack Overflow用户
发布于 2018-06-11 14:08:53
我的URL设置如下：
urlpatterns = [
url(r'^games/$', views.GameList.as_view()),
url(r'^games/(?P<pk>[0-9]+)/$', views.GameDetail.as_view()),
]改为：
from django.conf.urls import url
from games import views
urlpatterns = [
url(r'^games', views.GameList.as_view()),
url(r'^games/(?P<pk>[0-9]+)/$', views.GameDetail.as_view()),
]页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/-100005290
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