django rest framework 创建方法序列化器

Question

我有这段工作代码，我想知道是否可以将代码写得更短。仅仅因为我必须指定代理，所以我正在传递如下所示的每个和平数据。

class ListingSerializer(serializers.ModelSerializer):

    rooms = serializers.SerializerMethodField()

    def create(self, validated_data):
        agent = self.context.get("agent")
        listing = Listing.objects.create(
            agent=agent,
            title=validated_data["title"],
            description=validated_data["description"],
            floor=validated_data["floor"],
            floor_count=validated_data["floor_count"],
            price=validated_data["price"],
            street=validated_data["street"],
            house_no=validated_data["house_no"],
            door_no=validated_data["door_no"],
            city=validated_data["city"],
            country=validated_data["country"],
            postal_code=validated_data["postal_code"],
            tilstand_report=validated_data["tilstand_report"],
            water_consumption_report=validated_data["water_consumption_report"],
            energy_level_report=validated_data["energy_level_report"],
            property_tax_report=validated_data["property_tax_report"],
        )

        return listing

Answer 1

您只需将代理“注入”到 validated_data:

class ListingSerializer(serializers.ModelSerializer):
    rooms = serializers.SerializerMethodField()

    def create(self, validated_data):
        <strong>validated_data['agent'] = self.context['agent']</strong>
        return super().create(validated_data)

django rest framework 创建方法序列化器

django rest framework create method serializer

django

serialization

django-rest-framework

django-serializer