DropDownList中的合格的SelectedValue在Html.BeginForm（）在AS

这是我的查看代码：

@using(Html.BeginForm(new { SelectedId = /*SelectedValue of DropDown*/ })) {

 <fieldset>

     <dl>
       <dt>
           @Html.Label(Model.Category)
       </dt>
       <dd>
        @Html.DropDownListFor(model => Model.Category, CategoryList)
       </dd>
    </dl>

 </fieldset>
 <input type="submit" value="Search" />


}

如图所示代码我需要将通过dropdown选择的值在动作BeginForm() HTML辅助。 什么是您的建议？

Answer 1:

因为下拉列表由表示的形式被提交时所选择的值将被传递<select>元素。 你只需要调整您的视图模型，以便它有一个叫做财产SelectedId例如要在其中绑定的下拉列表：

@using(Html.BeginForm() )
{
    <fieldset>
        <dl>
            <dt>
                @Html.LabelFor(x => x.SelectedId)
            </dt>
           <dd>
                @Html.DropDownListFor(x => x.SelectedId, Model.CategoryList)
           </dd>
        </dl>
    </fieldset>

    <input type="submit" value="Search" />
}

这假定以下视图模型：

public class MyViewModel
{
    [DisplayName("Select a category")]
    public int SelectedId { get; set; }

    public IEnumerable<SelectListItem> CategoryList { get; set; }
}

将由控制器进行处理：

public ActionResult Index()
{
    var model = new MyViewModel();
    // TODO: this list probably comes from a repository or something
    model.CategoryList = new[]
    {
        new SelectListItem { Value = "1", Text = "category 1" },
        new SelectListItem { Value = "2", Text = "category 2" },
        new SelectListItem { Value = "3", Text = "category 3" },
    };
    return View(model);
}

[HttpPost]
public ActionResult Index(MyViewModel model)
{
    // here you will get the selected category id in model.SelectedId
    return Content("Thanks for selecting category id: " + model.SelectedId);
}

文章来源: Pass SelectedValue of DropDownList in Html.BeginForm() in ASP.NEt MVC 3