Ajax将数据发布到PHP

我有以下代码：

  <?php
        $query=mysqli_query($mysqli, "SELECT assignment,a_id FROM assignments WHERE groupid='".$groupid."'");
        $assignments = array(); // create empty assignments array

        while ($row=mysqli_fetch_array($query)){
            echo'<td class="columnname" id="'.$row['a_id'].'" contenteditable="true">'.$row['assignment'].'</td>';
            $assignments[$row['a_id']] = $row['assignment']; // add assignment to array

        }
        ?>

      <script>
$('.columnname').keyup(function() {
    delay(function(){

        var text= $(this).text();
        var id= $(this).attr('id')
        $.ajax({
            type:"POST",
            url:"updateassignment.php",
            data:{text:text, id:id},
            success:function(data){
                console.log('success bro!');
            }
        });
    }, 500 );
});

var delay = (function(){
    var timer = 0;
    return function(callback, ms){
        clearTimeout (timer);
        timer = setTimeout(callback, ms);
    };
})();
</script>  

我找到了答案

这是因为我在delay函数中定义了ajax，所以它不会引用columnname。解决这个问题的方法是在delay函数之外定义$（this）

<?php
include_once("config.php");

if (isset($_POST['id']) && isset($_POST['text'])) {

    $id=$_POST['id'];
    $text=$_POST['text'];

    $query=mysqli_query($mysqli, "UPDATE assignments SET assignment='$text' WHERE a_id='$id'")or die(mysqli_error($mysqli));


}


?>