Question

我经常需要使用其他信息扩展我的域模型，然后再使用WebAPI将其返回给客户端。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。我想出了类似的东西：

首先是SerializeObject
然后Parse
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;

我这是正确的方法吗？

Answer 1

JObject实现了IDictionary，因此你可以这样使用它。例如，

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

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

您还可以使用dynamic关键字

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

Answer 2

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);

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,}

Answer 3

这将有效：

var cycles = cycleSource.AllCycles();

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

var vm = new JArray();

foreach (var cycle in cycles)
{
    var cycleJson = JObject.FromObject(cycle);
    // extend cycleJson ......
    vm.Add(cycleJson);
}

return vm;

使用Json.NET将任何类型的对象转换为JObject

3 个答案: