使用 Json.NET 将任何类型的对象转换为 JObject

问题描述

在使用 WebAPI 将其返回给客户端之前，我经常需要使用附加信息扩展我的域模型.为了避免创建 ViewModel，我想我可以返回带有附加属性的 JObject.但是，我无法找到直接方法将任何类型的对象转换为 JObject，只需调用 Newtonsoft JSON 库即可.我想出了这样的事情:

I often need to extend my Domain model with additional info before returning it to the client with WebAPI. To avoid creation of ViewModel I thought I could return JObject with additional properties. I could not however find direct way to convert object of any type to JObject with single call to Newtonsoft JSON library. I came up with something like this:

1. 第一个 SerializeObject
2. 然后解析
3. 并扩展 JObject

例如:

var cycles = cycleSource.AllCycles();

var settings = new JsonSerializerSettings
{
    ContractResolver = new CamelCasePropertyNamesContractResolver()
};

var vm = new JArray();

foreach (var cycle in cycles)
{
    var cycleJson = JObject.Parse(JsonConvert.SerializeObject(cycle, settings));
    // extend cycleJson ......
    vm.Add(cycleJson);
}

return vm;

我这个方法正确吗?

推荐答案

JObject 实现了 IDictionary，所以你可以这样使用它.例如，

JObject implements IDictionary, so you can use it that way. For ex,

var cycleJson  = JObject.Parse(@"{""name"":""john""}");

//add surname
cycleJson["surname"] = "doe";

//add a complex object
cycleJson["complexObj"] = JObject.FromObject(new { id = 1, name = "test" });

所以最终的 json 将是

So the final json will be

{
  "name": "john",
  "surname": "doe",
  "complexObj": {
    "id": 1,
    "name": "test"
  }
}

你也可以使用动态关键字

dynamic cycleJson  = JObject.Parse(@"{""name"":""john""}");
cycleJson.surname = "doe";
cycleJson.complexObj = JObject.FromObject(new { id = 1, name = "test" });

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