Asp.net mvc 4 asp.net mvc模式弹出窗口

Asp.net mvc 4 asp.net mvc模式弹出窗口,asp.net-mvc-4,popup,Asp.net Mvc 4,Popup,我正在使用一个简单的模式弹出窗口。我的目标是将对话框的信息发送到另一个视图。 下面是我的索引视图 <button id="modal-opener">Open</button> <div id="dialog-modal"> @using (Ajax.BeginForm("Index",new AjaxOptions{UpdateTargetId = "ID",HttpMethod = "Post",OnSuccess = "onSuccess

我正在使用一个简单的模式弹出窗口。我的目标是将对话框的信息发送到另一个视图。 下面是我的索引视图

    <button id="modal-opener">Open</button>
<div id="dialog-modal">
    @using (Ajax.BeginForm("Index",new AjaxOptions{UpdateTargetId = "ID",HttpMethod = "Post",OnSuccess = "onSuccess"}))
    {
        <div>
            <fieldset>
                <legend>Acount Information</legend>
                <div id="editor-label">
                    @Html.LabelFor(a=>a.FirstName)
                </div>
                <div id="editor-field">
                    @Html.TextBoxFor(a=>a.FirstName)
                    @Html.ValidationMessageFor(a=>a.FirstName)
                </div>
                <div id="editor-label">
                    @Html.LabelFor(a=>a.LastName)
                </div>
                <div id="editor-field">
                    @Html.TextBoxFor(a=>a.LastName)
                    @Html.ValidationMessageFor(a=>a.LastName)
                </div>
                <p>
                    <input type="submit" value="submit"/>
                </p>
            </fieldset>

        </div>

    }
</div>

如果要执行完全重定向而不是部分回发,则需要更改此行:

@using (Ajax.BeginForm("Index",new AjaxOptions{UpdateTargetId = "ID",HttpMethod = "Post",OnSuccess = "onSuccess"}))
进入

如果只想部分回发到details视图,则将from索引上的操作更改为details

@using (Ajax.BeginForm("Details",new AjaxOptions{UpdateTargetId = "ID",HttpMethod = "Post",OnSuccess = "onSuccess"}))

HTH

那么问题出在哪里呢?问题如下,当我单击按钮弹出关闭,但浏览器无法直接查看详细信息时
@using (Ajax.BeginForm("Index",new AjaxOptions{UpdateTargetId = "ID",HttpMethod = "Post",OnSuccess = "onSuccess"}))
 @using(Html.BeginForm("Index"))
@using (Ajax.BeginForm("Details",new AjaxOptions{UpdateTargetId = "ID",HttpMethod = "Post",OnSuccess = "onSuccess"}))