asp.net-mvc-4 – 将模型从View传递给Controller

我想将模型传递给控制器​​的Post方法.调用方法时,它显示内容的空值和0的Id.理想情况下,它应该包含它显示的模型的值.

视图：

@model MvcApplication4.Models.WorldModel
@{
    ViewBag.Title = "Information";
    Layout = "~/Views/Shared/_Layout.cshtml";
}

@using (Html.BeginForm("Information", "World", FormMethod.Post))
{
    @Html.AntiForgeryToken()
    @Html.ValidationSummary(true)
    <br />
    @Html.DisplayFor(model => model.Id)
    <br />
    @Html.DisplayFor(model => model.Content)
    <br />
    <input type="submit" value="Next" />
}

控制器：单击提交按钮时调用的方法.

[HttpPost]
        [ValidateAntiForgeryToken]
        public ActionResult Information(WorldModel worldModel)
        {
            CandidateSession cs = (CandidateSession)Session["Can"];
            var can = cs.Candidates.Where(x => x.IsNameDispalyed == false);
            if (can.Count() > 0)
            {
                var can1 = can.First();
                can1.IsNameDispalyed = true;
                Session["Can"] = cs;
                return View(new WorldModel() { Id = can1.Id, Content = can1.Name });
            }
            return View(new WorldModel());
        }

模型：

public class WorldModel
    {
        public int Id { get; set; }

        public string Content { get; set; }
    }

您应该添加包含您的值的隐藏输入：

@Html.HiddenFor(model => model.Id)
@Html.HiddenFor(model => model.Content)

这是因为模型绑定器将搜索输入(如文本框或隐藏字段)以获取值并将其与模型属性(基于输入名称)相关联.使用DisplayFor不会创建任何输入,因此模型绑定器在您提交表单时无法找到您的值.