重新绑定在AJAX中单击事件后使用一次

Question

我在AJAX调用中遇到点击事件时遇到问题。我有AJAX调用嵌套，因为点击事件发生在直到第一次AJAX调用之前不存在的div。基本上我从数据库加载用户评论，然后在评论上有一个向上和向下的投票按钮。这些是第二个Ajax调用。你可以在这里看到JS：

//FIRST LOAD THE COMMENTS WITH THE FIRST AJAX CALL 
$('.comments_submit').each(function(){ 
    var comments_text = $(this).parent().find('textarea').val(); 
    var video_id = $(this).closest('div.home_video').find('.video').attr('id'); 

    var myData = 'body=' + comments_text + '&video_id=' + video_id + '&type=load'; 
    var that = this; 

    $.ajax({ 
       type: 'POST', // HTTP method POST or GET 
       url: 'comments_ajax.php', //Where to make Ajax calls 
       dataType:'text', // Data type, HTML, json etc. 
       data:myData, //post variables 
       success:function(response){ 

       //UPON SUCCESSFULLY LOADING THE COMMENTS/BIND CLICK EVENT TO VOTE UP AND VOTE DOWN 
        $(that).closest('div.home_video').find('.comments_list').html(response); 

        $('.vote_link.down').each(function(){ 
         $(this).click(function(){ 
          var comment_id = $(this).closest('.comment_box').attr('id'); 

          //MAKE AJAX CALL TO SUBMIT COMMENTS 

          var myData = 'comment_id=' + comment_id + '&type=vote_down'; 
          var that = this; 

          $.ajax({ 
           type: 'POST', // HTTP method POST or GET 
           url: 'comments_ajax.php', //Where to make Ajax calls 
           dataType:'text', // Data type, HTML, json etc. 
           data:myData, //post variables 
           success:function(response){ 

           //REFORMAT UPON SUCCESSFUL AJAX CALL 
           $(that).text(response); 

           }, 
           error:function (xhr, ajaxOptions, thrownError){ 
            alert('didnt work'); //throw any errors 
           } 
          }); 
         }); 
        }); 


        $('.vote_link.up').each(function(){ 
         $(this).click(function(){ 
          var comment_id = $(this).closest('.comment_box').attr('id'); 

          //MAKE AJAX CALL TO SUBMIT COMMENTS 

          var myData = 'comment_id=' + comment_id + '&type=vote_up'; 
          var that = this; 

          $.ajax({ 
           type: 'POST', // HTTP method POST or GET 
           url: 'comments_ajax.php', //Where to make Ajax calls 
           dataType:'text', // Data type, HTML, json etc. 
           data:myData, //post variables 
           success:function(response){ 

           //REFORMAT UPON SUCCESSFUL AJAX CALL 
           $(that).text(response); 

           }, 
           error:function (xhr, ajaxOptions, thrownError){ 
            alert('didnt work'); //throw any errors 
           } 
          }); 
         }); 
        }); 



       }, 
       error:function (xhr, ajaxOptions, thrownError){ 
        alert('didnt work'); //throw any errors 
       } 
     }); 
}); 

的评论负载的服务器端代码：

if($_POST['type']=="load"){ 
$sql_select = "SELECT * FROM comments WHERE video_id = '$video_id' ORDER BY timestamp DESC LIMIT 7"; 
$result_select = $mysqli->query($sql_select); 

while($row = mysqli_fetch_array($result_select)){ 
    echo "<div class='comment_box' id='".$row['id']."'>".$row['body']." 
     <div class='votes'> 
      <a class='vote_link up'>UP (".$row['vote_up'].")</a> 
      <a class='vote_link down'>DOWN (".$row['vote_down'].")</a> 
     </div> 
    </div>"; 
} 

}

而对于投票支持服务器端代码（这是相当多相同表决向下代码）：

if($_POST['type']=="vote_up"){ 
$comment_id = $_POST['comment_id']; 

$sql_select = "SELECT * FROM comments WHERE id = '$comment_id'"; 
$result_select = $mysqli->query($sql_select); 
$row = mysqli_fetch_array($result_select); 
$votes = $row['vote_up'] + 1; 

$sql_update = "UPDATE comments SET vote_up = '$votes' WHERE id = '$comment_id'"; 
$result_update = $mysqli->query($sql_update); 

$sql_select = "SELECT * FROM comments WHERE id = '$comment_id'"; 
$result_select = $mysqli->query($sql_select); 
$row_vote = mysqli_fetch_array($result_select); 

echo " 
UP (".$row_vote['vote_up'].") 
"; 

}

一对夫妇。当我点击这个按钮时，出于某种原因它会向UP或DOWN字段添加多张票。但它也会取消绑定点击事件。

我从来没有嵌套的AJAX调用之前，所以也许有绑定的AJAX功能的第一功能，将发生后更好的办法？谢谢你的帮助！

Answer 1

你应该能够做到：

$('.vote_link.down').click(function(){ 
    var comment_id = $(this).closest('.comment_box').attr('id'); 
    // etc. 
}); 

$('.vote_link.up').click(function(){ 
    var comment_id = $(this).closest('.comment_box').attr('id'); 
    // etc. 
}); 

这应该简化您的代码，并可能会解决您的一些问题。

Answer 2

我找到了答案（对于有同样问题的其他人）。而不是绑定顶级AJAX调用内的函数，只需使用$（document）.on（）来调用函数，如下所示：

$(document).on("click",".vote_link.down",function(e){ 
    alert('testing'); 
});