Django 不在URL中传递ID-UpdateAliView

我想使用我的

API

进行更新。为此，我在

URL

中传递

id

，它是

pk

http://localhost:8000/api/manager/update/96

其中96是

主键

。现在，我不想在url中传递

id

，而是想在正文中传递

id

，并更新数据。我的url应该是这样的

http://localhost:8000/api/manager/update

Views.py

class ManagerUpdateAPIView(APIView):

    def post(self, request, pk, *args, **kwrgs):

        user = get_object_or_404(User, id=pk)
        userprofile = get_object_or_404(UserProfile, user=pk)

        serializer1 = EmployeeRegisterSerializer(user, data=request.data)
        serializer2 = EmployeeProfileSerializer(userprofile, data=request.data)

        user_role = ACLRoles.objects.get(id=4)

        if serializer1.is_valid() and serializer2.is_valid():

            serializer1.save()
            serializer2.save()

            return Response(status=status.HTTP_200_OK)
        print(serializer1.errors)
        print(serializer2.errors)
        return Response(status=status.HTTP_404_NOT_FOUND)

序列化程序.py

class EmployeeProfileSerializer(serializers.ModelSerializer):
    class Meta:
        model = UserProfile
        fields = [  
                    'user_employee_id',
                    'user_phone', 
                    'user_payroll_id',
                    'user_hire_date',
                    'user_pay_rate',
                    'user_salaried',
                    'user_excempt',
                    'user_groups',
                    'user_state',
                    'user_city',
                    'user_zipcode', 
                    'user_status',
                ]

class EmployeeRegisterSerializer(serializers.ModelSerializer):
    # userprofile = EmployeeProfileSerializer(read_only=True)

    class Meta:
        model = User
        fields = ['first_name','last_name', 'email',]

如何更新数据而不在url中传递

id

。

请尝试此代码

class ManagerUpdateAPIView(APIView):

    def post(self, request, *args, **kwrgs): #change is here
        pk = request.data['pk'] #change is here
        user = get_object_or_404(User, id=pk)
        userprofile = get_object_or_404(UserProfile, user=pk)

        serializer1 = EmployeeRegisterSerializer(user, data=request.data)
        serializer2 = EmployeeProfileSerializer(userprofile, data=request.data)

        user_role = ACLRoles.objects.get(id=4)

        if serializer1.is_valid() and serializer2.is_valid():
            serializer1.save()
            serializer2.save()

            return Response(status=status.HTTP_200_OK)
        print(serializer1.errors)
        print(serializer2.errors)
        return Response(status=status.HTTP_404_NOT_FOUND)

---

如果我没有发送数据，其中一个字段会出现错误-

serializer2{'salaried'：[ErrorDetail（string='This field is required'，code='required'）]，'excempt'：[ErrorDetail（string='This field is required'，code='required'）]

您也必须在消息payload中传递这些数据。我不知道是哪个负载。我正在传递

serializer2=EmployeeProfileSerializer（userprofile，data=request.data）

Payload是您发送给drf的数据，我如何传递它，因为

serializer1=EmployeeRegisterSerializer（user，data=request.data）

即使我不发送任何数据，它也在完美地更新我的数据。Just

serializer2

不起作用

{
    "pk":96,
    "other_data":"other data of your usual payload"
}