Question

序列化器中的我的变量 MySerializer。

du = serializers.SerializerMethodField()

def get_du(self, obj):

   modelname = Somemodel.objects.filter('id=id')
   du = sum(modelname.values_list('t', flat=True))
   return du

我想按du的值以降序返回响应。我的看法：

    def get(self, request):
        queryset = Anothermodel.objects.all()
        serializer = MySerializer(queryset, many=True)
        return Response({"organization": serializer.data}, status=status.HTTP_200_OK)

Answer 1

设法通过帮助How to sort values of serializer custom field in DRF

解决了该问题

serializer_data = sorted(serializer.data, key=lambda k: k['du'], reverse=True)
        return Response({"info": serializer_data}, status=status.HTTP_200_OK)

Answer 2

您是否尝试过使用.order_by（）？像这样：

AnotherModel.objects.all().order_by('-du')

https://docs.djangoproject.com/en/3.0/ref/models/querysets/

在序列化程序自定义字段中订购

2 个答案: