Question

我具有如下所示的当前api，我的目的是使其显示所需的输出。我想将年龄和国籍（如下所示）（请参阅所需的输出）添加到嵌套的类型中，但是它似乎仅以不嵌套的纯格式显示。使其嵌套的最佳方法是什么？

  "type" : [
    { "age" : "27",  "nationality" : "usa" }

当前API

[
    {
     "name"    : "Jay",
     "school" : "college",
     "type" : "usa"
     "age": "27"
     "nationality": "usa"

        },
    ]

所需的API

[
    {
    "name"    : "Jay",
    "school" : "college",
    "type" : [
     { "age" : "27",  "nationality" : "usa" },
     { "age" : "24",  "nationality" : "canada" },
     { "age" : "26",  "nationality" : "thailand" },
  ]
}
]

Serializers.py

class SchoolSerializer(serializers.ModelSerializer):
    class Meta:
        model = School
        fields = ("name", "school", "type", "age", "nationality")

    def update(self, instance, validated_data):
        instance.name = validated_data.get("name", instance.name)
        instance.school = validated_data.get("school", instance.session)
        instance.type = validated_data.get("type", instance.session)
        instance.age = validated_data.get("age", instance.session)
        instance.nationality = validated_data.get("nationality", instance.session)


    instance.save()
    return instance

models.py

class School(models.Model):
    name = models.CharField(max_length=255, null=False)
    school = models.CharField(max_length=255, null=False)
    type = models.CharField(max_length=255, null=False)
    age = models.CharField(max_length=255, null=False)
    nationality = models.CharField(max_length=255, null=False)

    def __str__(self):
        return "{} - {}".format(self.name, self.school)

views.py

class ListSchoolView(generics.ListCreateAPIView):
    """
    Provides a get method handler.
    """
    queryset = School.objects.all()
    serializer_class = SchoolSerializer

Answer 1

如果要自定义模型的序列化方式，请覆盖序列化器的to_representation方法。这似乎与REST API设计背道而驰，因此我不建议这样做。您应该为TypeSerializer创建一个单独的序列化程序，以获得最佳结果。

将此方法添加到SchoolSerialzer中以实现所需的格式：

def to_representation(self, obj):
    rep = super().to_representation(obj)
    age = rep.pop('age')
    type_ = rep.pop('type')
    rep['type'] = {
        'age': age,
        'nationality': type_
    }
    return rep

Django REST嵌套对象序列化器

1 个答案: