使用外键向Django Rest Framework API发送修补程序请求

我需要将放置请求发送给我只知道外键的API。应该怎么做到这一点。使用外键向Django Rest Framework API发送修补程序请求

models.py

class Company(models.Model): 
    name = models.CharField(max_length = 100) 
    user = models.OneToOneField(settings.AUTH_USER_MODEL, unique = True)

serializer.py

class CompanySerializer(serializers.ModelSerializer): 

    class Meta: 
     model = Company 
     fields = ('id', 'name','user')

views.py

class Company(object): 
    permission_classes = (IsAuthenticated,IsUserThenReadPatch) 
    serializer_class = CompanySerializer 

    def get_queryset(self): 
     user = self.request.user 
     return Company.objects.filter(user = user) 

class CompanyDetails(Company, RetrieveUpdateAPIView, APIView): 

    pass

urls.py

url(r'^company/$',views.CompanyDetails.as_view()),

来源

2015-01-21 mithu

为了能够在DRF所有CRUD操作，你可能想使用ViewSet而不是View：

# views.py 
class CompanyViewSet(ViewSet): 
    permission_classes = (IsAuthenticated,IsUserThenReadPatch) 
    serializer_class = CompanySerializer 

    def get_queryset(self): 
     user = self.request.user 
     return Company.objects.filter(user = user) 

# urls.py 
router = routers.SimpleRouter() 
router.register(r'company', CompanyViewSet) 
urlpatterns = router.urls

以上将允许你这样做将所有CRUD REST请求：

GET /company - 列表中的所有公司
POST /company - 创建公司
GET /company/:id - 得到一个单一的公司
PUT/POST /company/:id - 更新公司
PATCH /company/:id - 部分更新公司
DELETE /company/:id - 删除公司

您可以在DRF文档阅读更多 - viewsets和routers

来源

2015-01-21 13:43:02 miki725

使用外键向Django Rest Framework API发送修补程序请求

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