Asp.net mvc 4 asp.net mvc模式弹出窗口
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
<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"}))