将模型字段序列化为嵌套对象/字典

Question

想象以下模型：

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

我有一个 django rest 框架ModelSerializer，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。所以当序列化为 json 输出时：

{
    name: 'Some name',
    address: {
        streetname: 'This is a test',
        housenumber: '23',
        zip: '1337',
    }
}

我尝试创建一个AddressSerializer

class Address(object):
    ...

class AddressSerializer(serializers.Serializer):
    streetname = serializers.CharField()
    housenumber = serializers.CharField()
    zip = serializers.CharField()
    ...

然后将PersonSerializer.address设置为使用AddressSerializer

class PersonSerializer(serializers.ModelSerializer):
    ...
    address = AddressSerializer()

这导致我的架构是正确的。我使用drf-yasg 生成swagger 文档。它查看序列化程序以生成正确的模型定义。所以序列化器需要表示模式。

这就是我目前所处的位置。显然现在它失败了，因为Person 模型中没有address 属性。你将如何解决这个问题？

Answer 1

来自DRF-doc for source 说，

值source='*'有特殊含义，用于表示 整个对象应该被传递到该字段。这个可以 对于创建嵌套表示或对于 需要访问完整的对象以确定输出 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

你可以在序列化器中定义一个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}