如何重新加載視圖後調用JQuery做動作和顯示消息

Question

我試圖找出一種方法來重新加載表後，每個按鈕的動作。我正在考慮重新加載頁面，但我不知道如何重新加載頁面，然後將從WebService函數返回的消息打印到div中。

CSHTML代碼：

@model Namespace.Models.ItemModel 
<div id="deleted" /> 
<input text id="curSelId" /> 
<input text id="curSelObj" /> 
<input text id="curSelfObjId" /> 
<table> 
<tbody> 
    @foreach (var item in Model.myDeleted) 
    { 
     <tr id="@item.DeletedId" value="@item.DeletedId"> 
      <td>@item.Timestamp</td> 
      <td class="type">@item.Type</td> 
      <td class="typeid">@item.TypeId</td> 
     </tr> 
    } 
</tbody> 
</table> 

JQuery的代碼：

function Undo() 
    { 
     $.ajax({ 
      type: "POST", 
      url: "@Url.Content("~/webservices/retrieve.asmx/Undo")", 
      data: "{'index': '" + $('[id$=curSelId]').val() + "'," + 
        "'type': '" + $('[id$=curSelObj]').val() + "'," + 
        " 'typeId': '" + $('[id$=curSelObjId]').val() + "'}", 
      contentType: "application/json; charset=utf-8", 
      dataType: "json", 
      success: function (msg) { 

       //http://encosia.com/2009/06/29/never-worry-about-asp-net-ajaxs-d-again/ 
       if (msg.hasOwnProperty("d")) { 
        // Leave the .d behind and pass the rest of 
        // the JSON object forward. 
        var obj = msg.d; 
        $("#deleted").html(obj); 
       } 
       else { 
        // No .d; no transformation necessary. 
        AjaxSucceeded(msg, dropdown, hidden); 
       } 
      } 
      , error: AjaxFailed 
     }); 
    } 

被修改：增加輸入字段。

Answer 1

一個重新加載它的方式是使用AJAX：

@model Namespace.Models.ItemModel 
<div id="deleted"></div> 
<div id="mytable" data-url="@Url.Action("MyTable", "SomeController")"> 
    @Html.Partial("_MyTable", Model) 
</div> 

其中_MyTable.cshtml包含表：

@model Namespace.Models.ItemModel 
<table> 
<tbody> 
    @foreach (var item in Model.myDeleted) 
    { 
     <tr id="@item.DeletedId" value="@item.DeletedId"> 
      <td>@item.Timestamp</td> 
      <td class="type">@item.Type</td> 
      <td class="typeid">@item.TypeId</td> 
     </tr> 
    } 
</tbody> 
</table> 

和你的Ajax請求的成功回調中重新加載DIV：

success: function (msg) { 
    var myTableDiv = $('#mytable'); 
    myTableDiv.load(myTableDiv.data('url')); 
    ... 
} 

其中MyTable動作會呈現相同的部分：

public ActionResult MyTable() 
{ 
    ItemModel model = ... 
    return PartialView("_MyTable");  
}