Question

我有一个带有mysql数据的表，我添加了一个垃圾按钮，我希望在使用ajax函数单击垃圾按钮时删除每一行，这是我的HTML：

  <table border="1">
    <?php
$sql ="SELECT * FROM music";
$result = mysql_query($sql) or die(mysql_error());

while($row = mysql_fetch_object($result)){

    echo '<tr><td>'.$row->file_name.'</td><td>'.$row->composer.'</td><td>'.$row->lyric.'</td><td>'.$row->music_thumb.'</td><td>'.'

    <a href="#" id="'.$row->msuic_id.'" class="trash" >
    جذف کردن
    </a>

    '.'</td></tr>';
    }

?>
  </table>

和我的ajax函数在这里：

$(function(){
        $('.trash').click(function(){
            var del_id= $(this).attr('id');
            var $ele = $(this).parent().parent();
            $.ajax({
                type:'POST',
                url:'delete.php',
                data:del_id,
                success: function(data){
                    if(data=="YES"){
                        $ele.fadeOut().remove();
                        }else{
                            alert("can't delete the row")
                            }
                    }

                })
            })
    });

以及我的“delete.php”页面：

<?php
include('../db_inc.php');
$music_number = "POST['del_id']";
echo '$music_number';
$qry = "DELETE FROM music WHERE msuic_id ='$music_number'";
$result=mysql_query($qry);

?>

我认为我的问题是ajax功能; 感谢

Answer 1

试试这个

$.ajax({
    type:'POST',
    url:'delete.php',
    data:{del_id:del_id},
    success: function(data){
         if(data=="YES"){
             $ele.fadeOut().remove();
         }else{
             alert("can't delete the row")
         }
    }

     })
})

并且还要改变

$music_number = "POST['del_id']";

到

$music_number = $_POST['del_id'];

Answer 2

你的ajax代码应该是这样的：

$(function(){
    $(document).on('click','.trash',function(){
        var del_id= $(this).attr('id');
        var $ele = $(this).parent().parent();
        $.ajax({
            type:'POST',
            url:'delete.php',
            data:{'del_id':del_id},
            success: function(data){
                 if(data=="YES"){
                    $ele.fadeOut().remove();
                 }else{
                        alert("can't delete the row")
                 }
             }

            });
        });
});

PHP代码应该是：

<?php
include('../db_inc.php');
$music_number = $_POST['del_id'];
//echo $music_number;
$qry = "DELETE FROM music WHERE msuic_id ='$music_number'";
$result=mysql_query($qry);
if(isset($result)) {
   echo "YES";
} else {
   echo "NO";
}
?>

Answer 3

除上述答案外，您还应委托点击处理程序以防止不必要的重复

$(document).on('click', '.trash', function() { ... });

Answer 4

将数据作为对象发送，而不仅仅是值

 ...
 type:'POST',
 url:'delete.php', 
 data:{'del_id':del_id},  //<----here
 ....

并在delete.php中将其作为POST

 $music_number = $_POST['del_id'];

<强>更新

将此添加到您的delete.php

<?php
 include('../db_inc.php');
 $music_number = "POST['del_id']";
  $qry = "DELETE FROM music WHERE msuic_id ='$music_number'";
  $result=mysql_query($qry);
  if($result) {
      echo "Yes";
   } else {
      echo "No";
   }
 ?>

Answer 5

试试这个：

$music_number = POST['del_id']; in delete.php

write ajax function like:

$.ajax({
                type:'POST',
                url:'delete.php',
                data:del_id,
                success: function(data){
                    if(data=="YES"){
                        $ele.fadeOut().remove();
                        }else{
                            alert("can't delete the row")
                            }
                    }

                })
            });

感谢

Answer 6

以下是您需要纠正的事项

在“delete.php”文件中 $music_number = "POST['del_id']"; // to $music_number = $_POST['del_id'];

此外，在ajax的成功回调中，您正在检查“YES”作为响应，而不是在此文件中的任何位置发送。
更改为您的ajax请求

data: {'del_id':del_id},

希望这有帮助。

用ajax函数和php删除行

6 个答案: