5

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ModelSerializer暴露所有字段的 django rest 框架。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：

{
    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()
    ...

然后设置PersonSerializer.address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. 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，因为模型中没有address属性Person。你将如何解决这个问题？

4

2 回答 2

11

来自DRF-docsource 说，

该值source='*'具有特殊含义，用于指示应将整个对象传递给该字段。这对于创建嵌套表示或需要访问完整对象以确定输出表示的字段很有用。


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
于 2018-06-08T10:08:21.887 回答
2

您可以property在序列化程序中定义 a ：

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}
于 2018-06-08T08:43:44.050 回答