MVC4：模型的属性为空 - 2

Question

由于某种原因，当我按下视图的任何按钮时，传递给操作方法的模型的所有属性都为空：

查看：

@using (Html.BeginForm("FolderChange", "EdiSender", FormMethod.Post, new {id = "ediFilesForm"}))
{
    var directoriesSelectList = new SelectList(Model.Directories);
    @Html.DropDownListFor(m => m.SelectedDirectory, directoriesSelectList, new {@Id = "Directories", @style = "width:Auto;", @size = 20, onchange = "$('#ediFilesForm').submit()", name = "action:FolderChange"})

    var ediFilesSelectList = new SelectList(Model.EdiFileNames);
    @Html.DropDownListFor(m => m.SelectedEdiFile, ediFilesSelectList, new {@Id = "EdiFileNames", @style = "width:Auto;", @size = 20})
}

<br/>
...

<form action="" method="post">
    <input type="submit" value="Send" name="action:Send" />
    <input type="submit" value="Delete" name="action:Delete" />
    <input type="submit" value="Refresh" name="action:Refresh" />
</form>

控制器：

    [HttpPost]
    [MultipleButton(Name = "action", Argument = "Send")]
    public ActionResult Send(EdiFileModel ediFileModel)
    {
           ....
    }

    [AttributeUsage(AttributeTargets.Method, AllowMultiple = false, Inherited = true)]
public class MultipleButtonAttribute : ActionNameSelectorAttribute
{
    public string Name { get; set; }
    public string Argument { get; set; }

    public override bool IsValidName(ControllerContext controllerContext, string actionName, MethodInfo methodInfo)
    {
        var isValidName = false;
        var keyValue = string.Format("{0}:{1}", Name, Argument);
        var value = controllerContext.Controller.ValueProvider.GetValue(keyValue);

        if (value != null)
        {
            controllerContext.Controller.ControllerContext.RouteData.Values[Name] = Argument;
            isValidName = true;
        }

        return isValidName;
    }
}

当我的按钮在 Html.BeginForm() 块中时它曾经可以工作，但我不能再这样了，因为它现在的操作名称为 FolderChange()，这与例如不同。 Send() 或其他处理按钮按下的操作方法。

谢谢。

已编辑：

@section scripts
{
    <sctipt type="text/javascript">
    $("#Directories").change(function () {
        var selectedDirectory = $("#Directories").val();   

        $(function () {
            $.getJSON('/DoWork/FolderChangeAjaxCall?selectedDirectory=' + selectedDirectory, function (result) {
                var ddl = $('#EdiFileNames');
                ddl.empty();
                $(result).each(function () {
                    $(document.createElement('option'))
                .attr('value', this.Id)
                .text(this.Value)
                .appendTo(ddl);
                });
            });
        });
    });    
    </sctipt>
}

<h2>Existing EDI Files</h2>
@using (Html.BeginForm("FolderChange", "EdiSender", FormMethod.Post, new {id = "ediFilesForm"}))
{
    var directoriesSelectList = new SelectList(Model.Directories);
    @Html.DropDownListFor(m => m.SelectedDirectory, directoriesSelectList, new {@Id = "Directories", @style = "width:Auto;", @size = 20})

    var ediFilesSelectList = new SelectList(Model.EdiFileNames);
    @Html.DropDownListFor(m => m.SelectedEdiFile, ediFilesSelectList, new {@Id = "EdiFileNames", @style = "width:Auto;", @size = 20})

Answer 1

在这里，它可能与您认为的答案相去甚远，但它会级联下拉菜单。您将不得不替换您的价值观和背景，但它会起作用。现在您可以将按钮放置在表单中，而不必让下拉菜单提交表单，这无论如何都是很糟糕的设计。

@Html.DropDownListFor(x => x.ddlID, Model.myDDLList, new { id = "ddl" }) 

@Html.DropDownListFor(x => x.myDDLList2, Model.myDDLList2, new { id = "foo" }) 

脚本

$("#ddl").change(function () {

        var Id = $("#ddl").val(); //1st dropdown Value        

        $(function () {
            $.getJSON('/DoWork/AjaxCall?Id=' + Id, function (result) {
                var ddl = $('#foo');
                ddl.empty();
                $(result).each(function () {
                    $(document.createElement('option'))
                .attr('value', this.Field1)
                .text(this.Field2)
                .appendTo(ddl);
                });
            });
        });
    });

控制器

public ActionResult AjaxCall(int Id)
    {
        using (PerformanceEntities context = new PerformanceEntities())
        {             
            return this.Json(
                  (from obj in context.ACCOUNT.Where(x => x.ACCT_ID == Id).ToList() 
                   select new
                   { 
                       Field1 = obj.ACCT_ID,
                       Field2 = obj.ACCT_NAME 
                   }),
                  JsonRequestBehavior.AllowGet
               );
        }         
    }

您可以查看this的帖子以供参考