Question

想象一下以下模型：

term_taxonomy_id

我有一个django休息框架class Person(models.Model): name = models.CharField() address_streetname = models.CharField() address_housenumber = models.CharField() address_zip = models.CharField()，它公开了所有字段。但我希望能够将地址字段序列化为字典。因此，当序列化为json输出时将是：

ModelSerializer

我尝试创建创建{ name: 'Some name', address: { streetname: 'This is a test', housenumber: '23', zip: '1337', } }

AddressSerializer

然后将class Address(object): ... class AddressSerializer(serializers.Serializer): streetname = serializers.CharField() housenumber = serializers.CharField() zip = serializers.CharField() ...设置为使用PersonSerializer.address

AddressSerializer

这导致我的架构正确。我使用class PersonSerializer(serializers.ModelSerializer): ... address = AddressSerializer()生成了一个swagger文档。它查看序列化程序以生成正确的模型定义。因此序列化程序需要表示模式。

所以这就是我现在所处的位置。显然现在它失败了，因为drf-yasg模型中没有address属性。你会怎么解决这个问题？

Answer 1

来自DRF-doc for source 的

表示，

值source='*'具有特殊含义，用于表示应该将整个对象传递给该字段。这个可以对于创建嵌套表示或用于创建嵌套表示的字段非常有用需要访问整个对象才能确定输出表示。

所以，试试这个，

class AddressSerializer(serializers.Serializer):
    streetname = serializers.CharField(source='address_streetname')
    housenumber = serializers.CharField(source='address_housenumber')
    zip = serializers.CharField(source='address_zip')


class PersonSerializer(serializers.ModelSerializer):
    # .... your fields
    address = AddressSerializer(source='*')

    class Meta:
        fields = ('address', 'other_fields')
        model = Person

Answer 2

您可以定义property：

class Person(models.Model):
    name = models.CharField()
    address_streetname = models.CharField()
    address_housenumber = models.CharField()
    address_zip = models.CharField()

    @property
    def address(self):
    return {'streetname': self.address_streetname,
            'housenumber': self.address_housenumber,
            'zip': self.address_zip}

将模型字段序列化为嵌套对象/字典

2 个答案: