ajax.reload第二次不工作

Question

我在mu MVC應用程序中使用數據表，我想每次刷新我的表，我通過模型彈出窗口編輯任何記錄。

它只適用於一個唱片，當我打開模型彈出第二次編輯相同/其他記錄它不。

請幫忙，這裏是我的代碼。

<div class="tablecontainer"> 
    <table id="schoolsTable" class="table table-bordered table-condensed table-hover table-striped"> 
     <thead> 
      <tr> 
       <th> 
        Status 
       </th> 
       <th> 
        Code 
       </th> 
       <th> 
        Name 
       </th> 
       <th> 
        Address 
       </th> 
       <th> 
        Actions 
       </th> 

      </tr> 
     </thead> 
    </table> 
</div> 
    <link href="//code.jquery.com/ui/1.11.4/themes/smoothness/jquery-ui.css" rel="stylesheet" /> 
    <link rel="stylesheet" href="https://cdnjs.cloudflare.com/ajax/libs/datatables/1.10.12/css/dataTables.bootstrap.min.css"> 
@section Scripts{ 
    <script src="~/Content/datatables/datatables.min.js"></script> 
    <script src="assets/lib/jquery/jquery.js"></script> 
    <script src="https://cdnjs.cloudflare.com/ajax/libs/jqueryui/1.12.0/jquery-ui.min.js"></script> 
    <script src="https://cdnjs.cloudflare.com/ajax/libs/datatables/1.10.12/js/jquery.dataTables.min.js"></script> 
    <script src="https://cdnjs.cloudflare.com/ajax/libs/datatables/1.10.12/js/dataTables.bootstrap.min.js"></script> 
    <script src="https://cdnjs.cloudflare.com/ajax/libs/jquery.tablesorter/2.26.6/js/jquery.tablesorter.min.js"></script> 


    <script> 
     $(document).ready(function() { 
      var oTable = $('#schoolsTable').DataTable({ 
       "ajax": { 
        "url": "/School/viewschool", 
        "type": "Get", 
        "datatype": "json" 
       }, 
       "columns": [ 
        { "data": "Mas_SchoolStatusID", "autowidth": true }, 
        { "data": "SchoolCode", "autowidth": true }, 
        { "data": "SchoolName", "autowidth": true }, 
        { "data": "SchoolAddress", "autowidth": true }, 
        { 
         "data": "Mas_SchoolID", "width": "50px", "render": function (data) { 
          return '<a class="popup" href=/School/Edit/' + data +'>Edit</a>'; 
         } 
        } 
       ] 
      }); 

      $('.tablecontainer').on('click', 'a.popup', function (e) { 
       e.preventDefault(); 
       OpenPopup($(this).attr('href')) 
      }) 

      function OpenPopup(pageUrl) 
      { 
       var $pageContent = $('<div/>'); 
       $pageContent.load(pageUrl); 
       $dialog = $('<div class="popupWindow" style="overflow:auto"></div>') 
         .html($pageContent) 
         .dialog({ 
          dragable: false, 
          autoOpen: false, 
          resizeable: false, 
          model: true, 
          title: 'Popup Dialog', 
          height: 550, 
          width: 600, 
          close: function() {         
           $dialog.dialog('distroy').remove(); 
          } 

         })     

       $('.popupWindow').on('submit', '#popupForm', function (e) { 
        var url = $('#popupForm')[0].action; 
        $.ajax({ 
         type: "POST", 
         url: url, 
         data: $('#popupForm').serialize(), 
         success: function (data) { 
          if (data.status) { 
           oTable.ajax.reload();       
           $dialog.dialog('close');        

          } 
         } 

        }) 
        e.preventDefault(); 
       }) 

       $dialog.dialog('open'); 
      } 

     }); 
    </script> 
    } 

Answer 1

我拼寫在破壞密切功能作爲distroy。

close: function() {         
        $dialog.dialog('distroy').remove(); 
        } 

這應該是：

close: function() {         
        $dialog.dialog('destroy').remove(); 
        }