Question

在我的控制器中，我有以下代码

public ActionResult 
{
    var roomdetails = db.RoomDetails.Include(r => r.RoomType).Include(r => r.FloorNames);

    roomdetails = roomdetails.OrderByDescending(s => s.FloorNames.FloorName);

    return View(roomdetails.ToList());
}

但我想发送roomdetails作为json对象，以便我可以使用jquery来捕获请求并在我的View中进行进一步的动态处理。所以如何将roomdetails转换为json对象。请帮助我......

Answer 1

注意：使用Newtonsoft.Json.dll版本8.0.2

将此类的实例返回给控制器方法。

public class ActionResult_Json : System.Web.Mvc.ActionResult
{
    public object To_Serialize_Object { get; set; }
    public ActionResult_Json(object To_Serialize_Object)
    {
        this.To_Serialize_Object = To_Serialize_Object;
    }
    public override void ExecuteResult(ControllerContext context)
    {
        context.HttpContext.Response.ContentType = "application/json";
        //serialize object to string
        string Serialized_Object_String =          Newtonsoft.Json.JsonConvert.SerializeObject(To_Serialize_Object, new Newtonsoft.Json.JsonSerializerSettings()
        {
            ReferenceLoopHandling = Newtonsoft.Json.ReferenceLoopHandling.Ignore,
        });
        //write json to response stream
        context.HttpContext.Response.Write(Serialized_Object_String);
    }
}

Answer 2

只需返回一个JsonResult：

public JsonResult Index()
{
    var roomdetails = db.RoomDetails.Include(r => r.RoomType).Include(r => r.FloorNames)
                                    .OrderByDescending(s => s.FloorNames.FloorName);
    return Json(roomdetails.ToList(), JsonRequestBehavior.AllowGet); 
}

你需要你的jquery以你想要的方式处理json并显示你想要的东西。通常情况下，如果通过ajax从您的视图中调用它，则效果很好。

Answer 3

@ {

ViewBag.Title = "Index";

}

接收

$（document）.ready（function（）{ $ .getJSON（'/ Reception / Index1'，function（data）{ 的console.log（数据）; if（data.success）{ 警报（ “辞海”）; // ShowStockQuote（数据）; } 其他{ 警报（ “faliure”）; } }）; }）;

@ Html.ActionLink（“创建新”，“创建”）

Answer 4

你可以试试这个。

public JsonResult Index()
{
    var roomdetails = db.RoomDetails.Include(r => r.RoomType).Include(r => r.FloorNames);
    roomdetails = roomdetails.OrderByDescending(s => s.FloorNames.FloorName);

    return Json(roomdetails.ToArray(), JsonRequestBehavior.AllowGet);
}

如何将LINQ结果转换为JSON并将其传递给MVC3中的View

4 个答案:

接收