最好的选择是从JsonResult类继承并覆盖Execute方法，例如

public class CustomJsonResult: JsonResult
{
    public CustomJsonResult()
    {
       JsonRequestBehavior = JsonRequestBehavior.DenyGet;
    }
    public override void ExecuteResult(ControllerContext context) {
            if (context == null) {
                throw new ArgumentNullException("context");
            }
            if (JsonRequestBehavior == JsonRequestBehavior.DenyGet &&
                String.Equals(context.HttpContext.Request.HttpMethod, "GET", StringComparison.OrdinalIgnoreCase)) {
                throw new InvalidOperationException(MvcResources.JsonRequest_GetNotAllowed);
            }

            HttpResponseBase response = context.HttpContext.Response;

            if (!String.IsNullOrEmpty(ContentType)) {
                response.ContentType = ContentType;
            }
            else {
                response.ContentType = "application/json";
            }
            if (ContentEncoding != null) {
                response.ContentEncoding = ContentEncoding;
            }
            if (Data != null) {
                CustomJsSerializer serializer = new CustomJsSerializer();
                response.Write(serializer.Serialize(Data));
            }
        }
}

代码来自mvc3的JsonResult类，并更改了此行

JavaScriptSerializer serializer = new JavaScriptSerializer();

至

CustomJsSerializer serializer = new CustomJsSerializer();

您可以在操作方法中使用此类，例如

public JsonResult result()
{
    var model = GetModel();
    return new CustomJsonResult{Data = model};
}

另外，您可以在Base控制器中覆盖Controller类的json方法，例如

public class BaseController:Controller
{
   protected internal override JsonResult Json(object data)
        {
            return new CustomJsonResult { Data = data };
        }
}

现在，如果您拥有BaseController中的所有控制器，return Json(data)则将调用序列化方案。Json您还可以选择重写方法的其他重载。