使用Json.NET将任何类型的对象转换为JObject

dra*_*fly 76 .net c# json.net

在将其返回到使用WebAPI的客户端之前,我经常需要使用其他信息扩展我的域模型.为了避免创建ViewModel,我想我可以使用其他属性返回JObject.然而,我无法通过单次调用Newtonsoft JSON库找到将任何类型的对象转换为JObject的直接方法.我提出了这样的事情:

  1. 首先是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;
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我这是正确的方法吗?

L.B*_*L.B 111

JObject实现了IDictionary,因此您可以这样使用它.对于前者,

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

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

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

{
  "name": "john",
  "surname": "doe",
  "complexObj": {
    "id": 1,
    "name": "test"
  }
}
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您也可以使用dynamic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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Con*_*dua 21

如果你有一个对象并希望成为JObject,你可以使用:

JObject o = (JObject)JToken.FromObject(miObjetoEspecial);
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像这样 :

Pocion pocionDeVida = new Pocion{
tipo = "vida",
duracion = 32,
};

JObject o = (JObject)JToken.FromObject(pocionDeVida);
Console.WriteLine(o.ToString());
// {"tipo": "vida", "duracion": 32,}
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  • 为什么不直接使用JObject.FromObject()代替(JObject)JToken.FromObject()? (9认同)