Question

我已将模型绑定到视图，模型中的每个属性在视图中都有一个关联的控件。

这是一个很大的模型，视图上有多个控件。用户更新视图中的数据，当他单击“保存”按钮时，应将更新的模型值发送到控制器操作方法。这里我正在对控制器方法进行AJAX调用。

由于这是强类型视图，我正在检查是否有可能直接传递更新的模型而不是访问控制值。

如果您需要任何进一步的信息，请与我们联系。我试过的PFB代码。

谢谢，巴拉斯

查看代码

    @model WebApplication1.Models.PersonModel

@{
    ViewBag.Title = "Index";
}

<h2>Index</h2>


@using (Html.BeginForm())
{
    @Html.AntiForgeryToken()

    <div class="form-horizontal">
        <h4>PersonModel</h4>
        <hr />
        @Html.ValidationSummary(true, "", new { @class = "text-danger" })
        <div class="form-group">
            @Html.LabelFor(model => model.Name, htmlAttributes: new { @class = "control-label col-md-2" })
            <div class="col-md-10">
                @Html.EditorFor(model => model.Name, new { htmlAttributes = new { @class = "form-control" } })
                @Html.ValidationMessageFor(model => model.Name, "", new { @class = "text-danger" })
            </div>
        </div>

        <div class="form-group">
            @Html.LabelFor(model => model.DateTime, htmlAttributes: new { @class = "control-label col-md-2" })
            <div class="col-md-10">
                @Html.EditorFor(model => model.DateTime, new { htmlAttributes = new { @class = "form-control" } })
                @Html.ValidationMessageFor(model => model.DateTime, "", new { @class = "text-danger" })
            </div>
        </div>

        <div class="form-group">
            @Html.LabelFor(model => model.Age, htmlAttributes: new { @class = "control-label col-md-2" })
            <div class="col-md-10">
                @Html.EditorFor(model => model.Age, new { htmlAttributes = new { @class = "form-control" } })
                @Html.ValidationMessageFor(model => model.Age, "", new { @class = "text-danger" })
            </div>
        </div>

        <div class="form-group">
            @Html.LabelFor(model => model.Company, htmlAttributes: new { @class = "control-label col-md-2" })
            <div class="col-md-10">
                @Html.EditorFor(model => model.Company, new { htmlAttributes = new { @class = "form-control" } })
                @Html.ValidationMessageFor(model => model.Company, "", new { @class = "text-danger" })
            </div>
        </div>

        <div class="form-group">
            <div class="col-md-offset-2 col-md-10">
                <input type="submit" value="Save" id="btnGet" class="btn btn-default" />
            </div>
        </div>
    </div>
}

<div>
    @Html.ActionLink("Back to List", "Index")
</div>

@section Scripts {
    @Scripts.Render("~/bundles/jqueryval")
}
<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.8.3/jquery.min.js"></script>
<script type="text/javascript">
    $(document).ready(function () {
        //window.setInterval(Setinterval, 10000);

            $("#btnGet").click(function () {
                Setinterval();
            });

            function Setinterval() {
                //var request = new PersonModel1();
                //var request = '<%= @Model %>';/// '<%= Model %>';
                var request = @Html.Raw(Json.Encode(Model));
                $.ajax({

                    url: "/ST/SubmitRequest",
                    dataType: "json",
                    contentType: "application/json",
                    type: "POST",
                    data: JSON.stringify(request),
                    success: function (response) {
                        //Setinterval();
                        alert("Done...!");
                    },
                    failure: function (response) {
                        alert(response.responseText);
                    },
                    error: function (response) {
                        alert(response.responseText);
                    }
                });
            };

        });


</script>

控制器代码

using System;
using System.Collections.Generic;
using System.Linq;
using System.Web;
using System.Web.Mvc;
using WebApplication1.Models;

namespace WebApplication1.Controllers
{
    public class STController : Controller
    {
        // GET: ST
        public ActionResult Index()
        {
            PersonModel pm = new PersonModel() {
                Age = "34",
                Company = "DDDD",
                DateTime = DateTime.Now,
                Name = "XYZ S"
            };
            return View(pm);
        }

        [HttpPost]
        public JsonResult SubmitRequest(PersonModel pm)
        {

            return Json(pm);
        }
    }
}

Answer 1

您正在使用（Html.BeginForm（）），因此您可以使用此代码

有一个功能正是这样做的：

http://api.jquery.com/serialize/

var data = $('form').serialize();
$.post('url', data);

在Jquery中访问强类型的ASP.Net MVC模型值

1 个答案: