我有一个动作,它返回一个特定类的对象的JsonResult.我已经使用一些attrib来装饰这个类的属性以避免使用null字段.类定义是：
private class GanttEvent
{
public String name { get; set; }
[JsonProperty(NullValueHandling = NullValueHandling.Ignore)]
public String desc { get; set; }
[JsonProperty(NullValueHandling = NullValueHandling.Ignore)]
public List<GanttValue> values { get; set; }
}
在我的动作中我使用了一个对象
var res = new List<GanttEvent>();
我回来使用：
return Json(res,JsonRequestBehavior.AllowGet);
不幸的是,我仍然在输出时收到空值：
[{"name":"1.1 PREVIOS AL INICIO ","desc":null,"values":null},{"name":"F04-PGA-S10","desc":"Acta preconstrucción",{"name":"F37-PGA-S10","desc":"Plan de inversión del anticipo",{"name":"F09-PGA-S10","desc":"Acta de vecindad",{"name":"F05-PGA-S10","desc":"Acta de inicio",{"name":"F01-PGA-S10","desc":"Desembolso de anticipo","values":null}]
我错过了什么或做错了什么？
解决方法
正如Brad Christie所说,MVC4仍然使用JavaScriptSerializer,因此为了让你的对象被Json.Net序列化,你将不得不执行几个步骤.
首先,从JsonResult继承一个新类JsonNetResult,如下所示(基于this solution)：
public class JsonNetResult : JsonResult
{
public JsonNetResult()
{
this.ContentType = "application/json";
}
public JsonNetResult(object data,string contentType,Encoding contentEncoding,JsonRequestBehavior jsonRequestBehavior)
{
this.ContentEncoding = contentEncoding;
this.ContentType = !string.IsNullOrWhiteSpace(contentType) ? contentType : "application/json";
this.Data = data;
this.JsonRequestBehavior = jsonRequestBehavior;
}
public override void ExecuteResult(ControllerContext context)
{
if (context == null)
throw new ArgumentNullException("context");
var response = context.HttpContext.Response;
response.ContentType = !String.IsNullOrEmpty(ContentType) ? ContentType : "application/json";
if (ContentEncoding != null)
response.ContentEncoding = ContentEncoding;
if (Data == null)
return;
// If you need special handling,you can call another form of SerializeObject below
var serializedObject = JsonConvert.SerializeObject(Data,Formatting.None);
response.Write(serializedObject);
}
}
然后,在您的控制器中,重写Json方法以使用新类：
protected override JsonResult Json(object data,JsonRequestBehavior behavior)
{
return new JsonNetResult(data,contentType,contentEncoding,behavior);
}
