显示在表Ajax响应的工作不

我有这样的剧本，我不知道我在这里失踪：显示在表Ajax响应的工作不

$(document).on('click','.btn-start-timer',function(){ 
      var id = $(this).val(), 
       time = $(this).parent().parent().find('.rendered_time').text(); 
       //console.log(time) 
       var i = window.setInterval(function(){ 

         var future = addMinutes(dat, 20); 
         //var future = new Date("Dec 12 2013 22:10:00 GMT-0800 (Pacific Standard Time) "); 
         var now = new Date(); 
         //console.log(tim); 
         var difference = Math.floor((future.getTime() - now.getTime())/1000); 

         var seconds = fixIntegers(difference % 60); 
         difference = Math.floor(difference/60); 

         var minutes = fixIntegers(difference % 60); 
         difference = Math.floor(difference/60); 

         var hours = fixIntegers(difference % 24); 
         difference = Math.floor(difference/24); 

         var days = difference; 

         $(this).parents('tr:first').find('td .days').html(seconds + "s"); 
         $(this).parents('tr:first').find('td .hours').html(minutes + "m"); 
         $(this).parents('tr:first').find('td .minutes').html(hours + "h"); 
         $(this).parents('tr:first').find('td .seconds').html(days + "d"); 

       if(hours == 0 && minutes == 0 && seconds == 0){ 
         window.clearInterval(i); 
         alert('times up'); 
        } 
       }, 1000); 

       });

，这里是我的html：

    <thead> 
         <tr> 
         <th>Client Name</th> 
         <th>Service Name</th> 
         <th>Time</th> 
         <th>Therapist Name</th> 
         <th>Status</th> 
         <th>Actions</th> 
         <th>Time</th>     
         </tr> 
        </thead> 
        <tbody> 


       <? $res = mysql_query("select * from rendered_services"); 
       while($row2 = mysql_fetch_array($res)){?> 
       <? $id_no = $row2['t_id']; 
       $selectstatus=$row2['therapist_status']; 

       ?> 


      <tr> 
      <td align="left"><?php echo $row2['client_name']; ?></td> 
      <td align="left"><?php echo $row2['rendered_service']; ?></td> 
      <td align="left" class="rendered_time"><?php echo $row2['rendered_time']?></td> 
      <td align="left"><?php echo $row2['assign_therapist']?></td> 
      <td align="left"><?php echo $row2['therapist_status']?></td> 

      <td align="left"> 
      <button class="btn btn-success btn-small btn-start-timer" value="<?php echo $row2['t_id']; ?>">Start</button> 
      <a class="btn btn-success btn-small" data-toggle="modal" href="pay_bill.php<?php echo '?t_id='.$id_no; ?>">Pay Bill</a> 
      </td> 
      <td> 
       <span class="days">a</span> 
       <span class="hours">a</span> 
       <span class="minutes">qa</span> 
       <span class="seconds">a</span> 
      </td> 
      </tr> 

      <?php }?> 
        </tbody> 
       </table>

我的问题是，当我单击.btn-start-timer时，数据或时间未显示在分配的选择器中。但是，当我把$(this).parents('tr:first').find('td .seconds').html(days + "d");例如setInterval之外它的工作！那么我在这里错过了什么？

来源

2013-12-13 leonardeveloper

见http://stackoverflow.com/questions/20279484/how-到接入的，正确的，这-上下文内，一个回调。 –

你正在失去setInterval功能

为了解决内部的this的情况下，宣布$(this)可变外setInterval

var $this=$(this) 
var id = $this.val(), 
time = $this.parent().parent().find('.rendered_time').text(); 

var i = window.setInterval(function(){ 
    /* replace instenaces of $(this) with $this*/ 
})

来源

2013-12-13 01:40:38 charlietfl

显示在表Ajax响应的工作不

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