如何在ListAPIView中使用django-filter对过滤结果进行排序

Posted 2021-04-25

我有一个ListAPIView使用DjangoFilterBackend根据url参数过滤房间。下面的代码就可以了。

现在，我想根据从Room对象的其他属性计算的分数，另一个url参数以及我们对发出请求的用户的了解来对结果进行排序。功能本身并不重要。

在应用我已有的过滤器后，如何对结果进行排序？

如果我自己进行过滤，我想我可以进行过滤，计算得分并在get_queryset中对结果进行排序，但在使用django-filter过滤后我不知道如何做到这一点。

示例查询

例如，我会执行此查询以按低于100的价格进行过滤.other_field值将用于计算排序分数：

http://localhost:8000/api/search/rooms?price=100&other_field=200

码

class RoomFilter(filters.FilterSet):
    price = filters.NumberFilter(name="price", lookup_expr='lte')
    features = filters.ModelMultipleChoiceFilter(
        name="features", 
        conjoined=True,
        queryset=Features.objects.all()
        )

    class Meta:
        model = Room
        fields = ['price', 'features']

class RoomSearchView(generics.ListAPIView):
    queryset = Room.objects.all()
    serializer_class = RoomSearchSerializer
    filter_backends = (filters.DjangoFilterBackend,)
    filter_class = RoomFilter

答案

尝试在API中覆盖list()，如下所示，

class RoomSearchView(generics.ListAPIView):
    queryset = Room.objects.all()
    serializer_class = RoomSearchSerializer
    filter_backends = (filters.DjangoFilterBackend,)
    filter_class = RoomFilter

    def list(self, request, *args, **kwargs):
        queryset = self.filter_queryset(self.get_queryset())

        queryset = queryset.order_by('-id')  # change is here  >> sorted with reverse order of 'id'

        page = self.paginate_queryset(queryset)
        if page is not None:
            serializer = self.get_serializer(page, many=True)
            return self.get_paginated_response(serializer.data)

        serializer = self.get_serializer(queryset, many=True)
        return Response(serializer.data)