Jquery：不推薦使用屬性的指定屬性。它總是返回true

Question

我有選擇的元素集合，當我改變了第一次一個選擇元素的值這個功能不工作，它的工作原理我第二次更改值和控制檯我有

jQuery代碼：

$(document).ready(function() { 
    var collection = $('select.ddlJ'); 
    console.log(collection); 
    for (var element in collection) { 
     $(element).change(function() { 

      $('select.ddlJ').change(function (e) { 
       $(this).parent().parent().find('td:last').prev().find('span').html(
        $(this).parent().parent().find('select.ddlJ').filter(function() { 
         return $.trim($(this).val()) == 'm'; 
        }).length); 

       $(this).parent().parent().find('td:last span').html(
       $(this).parent().parent().find('select.ddlJ').filter(function() { 
        return $.trim($(this).val()) == 'n'; }).length); 
      }); 
     }); 
    } 
}); 

HTML代碼：

<table cellspacing="0" border="1" style="border-collapse:collapse;" id="grid1" 
    rules="all"> 
     <tbody><tr> 
<th scope="col">Nr de ord</th> 
<th scope="col">StudentId</th> 
<th scope="col">Name LName</th> 

<th scope="col"> 
     <span id="mondayText">monday<br></span> 
     <span id="monday">14.05.2012</span> 
    </th> 
<th scope="col"> 
    <span id="thuesdayText">thuesday<br></span> 
    <span id="thuesday">15.05.2012</span> 
</th> 
<th scope="col"> 
    <span id="wednesdayText">wednesday<br></span> 
    <span id="wednesday">16.05.2012</span> 
</th> 
<th scope="col"> 
    <span id="thursdayText">thursday<br></span> 
    <span id="thursday">17.05.2012</span> 
</th> 
<th scope="col"> 
    <span id="fridayText">friday<br></span> 
    <span id="friday">18.05.2012</span> 
</th> 
<th scope="col"> 
    <span id="saturdayText">saturday<br></span> 
    <span id="saturday">19.05.2012</span> 
    </th> 
    <th scope="col"> 
    <span id="M">Total1</span> 
    </th> 
    <th scope="col"> 
    <span id="N">Total2</span> 
    </th> 
       </tr><tr> 
       <td> 1 </td> 
       <td>110001</td> 
       <td>Test1 Test1</td><td> 
       </td><td> 
       </td><td> 
       </td><td> 
<select class="ddlJ" id="a1111_0" name="ctl00$contentbody$grid1$ctl02$a1111"> 
       <option value="a" selected="selected">a</option> 
       <option value="m">m</option> 
       <option value="n">n</option> 
      </select> 
<select class="ddlJ" id="a2222_0" name="ctl00$contentbody$grid1$ctl02$a2222"> 
       <option value="a" selected="selected">a</option> 
       <option value="m">m</option> 
       <option value="n">n</option> 
      </select> 

          </td><td> 
          </td><td> 
          </td><td> 
          <span class="label" id="totalM"></span> 
          </td><td> 
          <span id="totalN"></span> 
          </td> 
     </tr><tr> 
     <td> 2 </td> 
       <td>110002</td> 
       <td>Test2 Test2</td> 
       <td></td> 
       <td></td> 
       <td></td><td> 
<select class="ddlJ" id="a1111_1" name="ctl00$contentbody$grid1$ctl03$a1111"> 
       <option value="a" selected="selected">a</option> 
       <option value="m">m</option> 
       <option value="n">n</option> 
      </select> 

       </td><td> 
       </td><td> 
       </td><td> 
       <span class="label" id="totalM"></span> 
       </td><td> 
       <span id="totalN"></span> 
        </td> 
     </tr><tr> 
      <td> 
         3     
       </td><td>110008</td><td>Test3 Test3</td><td> 
       </td><td> 
       </td><td> 
       </td><td> 
<select class="ddlJ" id="a1111_2" name="ctl00$contentbody$grid1$ctl04$a1111"> 
       <option value="a" selected="selected">a</option> 
       <option value="m">m</option> 
       <option value="n">n</option> 
      </select> 
         </td><td> 
       </td><td> 
       </td><td> 
       <span class="label" id="totalM"></span> 
       </td><td> 
        <span id="totalN"></span> 
          </td> 
     </tr> 
    </tbody> 
</table> 

Answer 1

這是jQuery中的一個錯誤： http://bugs.jquery.com/ticket/11397

（參見：https://stackoverflow.com/a/9460682/1178235）

Answer 2

嘗試改變

var collection = $('select.ddlJ') 

使用jQuery.get()

var collection = $('select.ddlJ').get();  
// BTW, this is similar to $('select.ddlJ')[0] 

，將給你實際的DOM元素，而不是一個jQuery對象匹配他們。或者（和我可能會做的方式），將代替改變

for (var element in collection) 

使用jQuery.each()

$('select.ddlJ').each(function(i, element) { ... }); 

Answer 3

嘗試改變.VAL（）來.attr（ '值'）