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Python Django Rest框架中的GET请求无法与响应JSON对象中的筛选器一起使用_Python_Json_Django_Django Views_Django Rest Framework - Fatal编程技术网

Python Django Rest框架中的GET请求无法与响应JSON对象中的筛选器一起使用

Python Django Rest框架中的GET请求无法与响应JSON对象中的筛选器一起使用,python,json,django,django-views,django-rest-framework,Python,Json,Django,Django Views,Django Rest Framework,Views.py 类用户点(ListAPIView): 或 序列化程序.py 类UserPointsSerializer(serializers.ModelSerializer): username=serializers.Field(source='User') 当我以用户名作为参数发送get请求时,我想获取与特定用户关联的点。您需要调用。在序列化程序上是否有效() def get(self, request, pk, format=None): user_attributes = U

Views.py

类用户点(ListAPIView):

序列化程序.py

类UserPointsSerializer(serializers.ModelSerializer): username=serializers.Field(source='User')


当我以用户名作为参数发送get请求时,我想获取与特定用户关联的点。

您需要调用。
在序列化程序上是否有效()

def get(self, request, pk, format=None):
    user_attributes = UserAttribute.objects.get(id=id)
    serializer = UserPointsSerializer(data=user_attributes)
    if serializer.is_valid():
        return Response(serializer.data)

但是您不能使用
is\u valid()
除非指定数据选项,否则必须是
用户点序列化器(data=user\u attributes)
此代码是一个语法错误
UserAttributes=self.get\u对象(id)
您不必将参数id放入self.get\u对象()因为它已经返回了具有请求id的对象实例,所以只能使用
self.get\u object()
。我建议您使用
retrieve(self,request,pk=None)
方法,而不是
get(self,request):
如果您只调用一个实例。
def get(self, request, pk, format=None):
    UserAttributes = self.get_object(id)
    serializer = UserPointsSerializer(UserAttributes)
    return Response(serializer.data)
class Meta:
    model = UserAttributes
    fields = [
        'points',
    ]
def get(self, request, pk, format=None):
    user_attributes = UserAttribute.objects.get(id=id)
    serializer = UserPointsSerializer(data=user_attributes)
    if serializer.is_valid():
        return Response(serializer.data)