Newtonsoft.Json C# :: 格式化 JsonConvert.SerializeObject

Question

我有以下课程，

  public class Actions
        {
            public string say { get; set; }
            public bool? listen { get; set; }
        }

        public class ActionsWrapper 
        { 
            public List<Actions> actions { get; set; }
            public ActionsWrapper(string say, bool listen = false)
            {
                this.actions = new List<Actions>();

                var action = new Actions();
                action.say = say;
                action.listen = listen;
                this.actions.Add(action);
            }
        }

我正在使用以下内容生成 Json

var actions = new ActionsWrapper(say: "Hi, how can I help you today?");
return JsonConvert.SerializeObject(actions);

这让我跟随 Json，

{"actions":[
      {
        "say":"Hi, how can I help you today?",
         "listen": false
       }
]}

这很好，但我将其发送到 Twilio API，它具有以下格式的要求，

{
  "actions": [
    {
      "say": "Hi, how can I help you today?"
    },
    {
      "listen": false
    }
  ]
}

所以我的问题是，我需要在我的类/NewtonSoft 中进行哪些更改才能在单独的花括号中获取每个属性 [say&listen]？

Answer 1

由于您的类已经被称为Actions ，您可以执行以下操作：

[Serializable]
public class Actions : ISerializable
{
    public string say { get; set; }
    public bool? listen { get; set; }

    public void GetObjectData(SerializationInfo info, StreamingContext context)
    {
        info.AddValue("actions", new object[] { new { say }, new { listen } });
    }
}

用法：

var actions = new Actions();
actions.say = say;
actions.listen = listen;
var json = JsonConvert.SerializeObject(actions);

Answer 2

使用 Newtonsoft.Json 的解决方案：

public class Say
{
    public string say { get; set; }
}

public class Listen
{
    public bool? listen { get; set; }
}

public class ActionsWrapper
{
    public List<Say> Says { get; set; }
    public List<Listen> Listens { get; set; }
    public ActionsWrapper(string say, bool listen = false)
    {
        this.Says = new List<Say>();
        this.Listens = new List<Listen>();
        Says.Add(new Say() { say = say });
        Listens.Add(new Listen() { listen = listen });
    }
}

用法：

var actions = new ActionsWrapper(say: "Hi, how can I help you today?");

JArray JArraySays = JArray.FromObject(actions.Says);
JArray JArrayListens = JArray.FromObject(actions.Listens);
JArraySays.Merge(JArrayListens);

return JsonConvert.SerializeObject(new { actions = JArraySays });