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来扩展我的域模型。为了避免创建ViewModel,我想我可以返回带有附加属性的JObject。然而,我找不到直接的方法来将任何类型的对象转换为JObject,只需调用一次Newtonsoft JSON库。我想出了这样的东西:
解析第一个SerializeObject
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;
我这条路走对了吗?
发布于 2014-02-24 22:59:39
JObject实现了IDictionary,所以您可以这样使用它。对于ex来说,
var cycleJson = JObject.Parse(@"{""name"":""john""}");
//add surname
cycleJson["surname"] = "doe";
//add a complex object
cycleJson["complexObj"] = JObject.FromObject(new { id = 1, name = "test" });
所以最后的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" });
发布于 2017-07-12 05:01:13
如果你有一个对象,并且希望成为JObject,你可以使用:
JObject o = (JObject)JToken.FromObject(miObjetoEspecial);
如下所示:
Pocion pocionDeVida = new Pocion{
tipo = "vida",
duracion = 32,
};
JObject o = (JObject)JToken.FromObject(pocionDeVida);
Console.WriteLine(o.ToString());
// {"tipo": "vida", "duracion": 32,}
发布于 2018-06-14 00:06:29
这将会起作用:
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;
https://stackoverflow.com/questions/21991223
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