Django 将项目添加到ListAPIView

我有一个普通的ListAPIView，它具有指定的模型和序列化程序类。我需要在列表的末尾再加一项。此项应手动创建（无查询集），并应具有与其他项相同的结构。

您可以覆盖此项的

列表

方法：

from rest_framework.response import Response

class YoourView(generics.ListAPIView):

    def list(self, request):
        queryset = self.get_queryset()
        serializer = UserSerializer(queryset, many=True)

        # append serializer's data with some additional value
        response_list = serializer.data 
        response_list.append(some_value)
        return Response(response_list)

试试这个，

from rest_framework import generics
from rest_framework.response import Response


class MyListView(generics.ListAPIView):
    serializer_class = MySerializer
    queryset = MyModel.objects.all()

    def list(self, request, *args, **kwargs):
        response_list = list(super().list(request, *args, **kwargs))
        custom_dict = {
            "key_1": "value 1",
            "key_2": "value 2"
        }
        response_list.append(custom_dict)
        return Response(data=response_list)

---

我认为最好保留

get\u queryset

、

filter\u queryset

和

get\u serializer

函数，以防修改

serializer\u类

，而不是使用默认的

serializer\u类

class YourView(ListAPIView):
    queryset = DefaultQueryset
    serializer_class = DefaultSerializer

    def list(self, request, *args, **kwargs):
        queryset = self.filter_queryset(self.get_queryset())
        serializer = self.get_serializer(queryset, many=True)
        response_list = serializer.data 
        response_list.append(some_value)
        return Response(response_list)

---

我建议您将实际负载与试图传递的额外数据分开。这将帮助您在稍后迭代这些冲突时避免很多冲突

试试这个：

class YourView(ListAPIView):
    queryset = YourModel.objects.all()
    serializer_class = YourSerializer

def list(self, request, args, **kwargs):
    serializer = self.get_serializer(self.get_queryset(), many=True)
    load = serializer.data
    newdict = {'your_new_data': 666}
    response_list = {'load': load, 'extra': newdict}
    return Response(response_list)