Question

对于具有字段“名称”的模型，我有一个简单的Django Rest Framework ModelSerializer。我想序列化它，所以名称是值，其清除的名称是按以下方式的键：

List<FileConfigData> configList = new List<FileConfigData>();
foreach (var file in fileConfigs){
   var configData = JsonConvert.DeserializeObject<FileConfigData>(File.ReadAllText(CONFIG_FILE_PATH)));
configList.Add(configData);
 }

我目前有：

"results": [
    {
        "mymodel1" : "My Model 1"
    },
    {
        "mymodel2" : "My Model 2"
    }]

返回：

class ModelSimpleSerializer(serializers.ModelSerializer):

  keyname = serializers.SerializerMethodField('get_model_keyname')
  class Meta:
    model = myModel
    fields = (
      'keyname',
    )
  def get_model_keyname(self,obj):
    keyname = obj.name
    keyname = keyname .lower()
    keyname = keyname .replace(" ", "")
    return keyname

有什么想法吗？

谢谢！

Answer 1

尝试一下

class ModelSimpleSerializer(serializers.Serializer):
    def to_representation(self, obj):
        keyname = obj.name.lower().replace(" ", "")
        return {
            keyname: obj.name,
        }

请注意，这将是只读的序列化程序。但这会满足您的要求

objects = myModel.objects.all()

ModelSimpleSerializer(objects.first()).data
# {"mymodel1" : "My Model 1"}

ModelSimpleSerializer(objects, many=True).data
# [{"mymodel1" : "My Model 1"}, {"mymodel2" : "My Model 2"}]

Answer 2

您可以尝试一下

class ModelSimpleSerializer(serializers.ModelSerializer):

      name = serializers.SerializerMethodField()
      class Meta:
        model = myModel
        fields = (
          'name',
        )
      def get_name(self,obj):
        keyname = obj.name
        keyname = keyname .lower()
        keyname = keyname .replace(" ", "")
        return keyname

使用功能更改ModelSerializer字段键

2 个答案: