如何为 Serializer 的查询集分页

Question

我正在检索 Category 及其 outfits 列表。我的问题是有太多 outfits 属于 category.

class CategoryListAPIView(generics.RetrieveAPIView):
    serializer_class = CategoryDetailSerializer
    ...

class CategoryDetailSerializer(serializers.ModelSerializer):
    outfits = serializers.SerializerMethodField()
    ...

    class Meta:
        model = Category
        fields = (
            ...
            'outfits',
            ...
        )

    def get_outfits(self, obj):  //This is returning 39 items. 
        // Can we paginate this? 
        if obj.outfits is not None:
            return OutfitListSerializer(obj.outfits, many=True).data
        return None

我们能否对其进行分页，以便用户可以首先看到 24 outfits 并刷新以查看其余 outfits？

Answer 1

如果你想要简单的条件 "first 24" 和 "the rest"。您可以通过获取参数来控制它。

def get_outfits(self, obj):
    show_all = self.request.GET.get('show_all')

    if show_all:
        outfits = obj.outfits.all()
    else:
        outfits = obj.outfits.all()[:24]

    return OutfitListSerializer(outfits, many=True).data

现在您可以使用 GET /categories/ 获取前 24 套服装的类别，使用 GET /categories/?show_all=true 获取完整展示

如何为 Serializer 的查询集分页

How to paginate queryset for Serializer

django

django-forms

django-pagination

django-serializer

django-rest-framework