MVC4 Action返回没有null的JsonResult

Far*_*lop 8 asp.net-mvc json asp.net-mvc-4

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; }
    }
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在我的Action中我使用了一个对象

    var res = new List<GanttEvent>();
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我回来使用:

    return Json(res, JsonRequestBehavior.AllowGet);
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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","values":null},{"name":"F37-PGA-S10","desc":"Plan de inversión del anticipo","values":null},{"name":"F09-PGA-S10","desc":"Acta de vecindad","values":null},{"name":"F05-PGA-S10","desc":"Acta de inicio","values":null},{"name":"F01-PGA-S10","desc":"Desembolso de anticipo","values":null}]
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我错过了什么或做错了什么?

Far*_*lop 8

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 此解决方案):

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);
    }
}
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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, string contentType, Encoding contentEncoding, JsonRequestBehavior behavior)
{
    return new JsonNetResult(data, contentType, contentEncoding, behavior);
}
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