javascript - 经管理员批准后更新数据库

Question

`我有一个显示推荐的网页。我只想显示数据库中“状态”为“1”的那些推荐。在管理员单击更新后，如何立即将名为“status”的数据库列从“0”更新为“1”。我也在使用 AJAX。

<tr align='center'>
<?php
include ('includes/connect.php');
$query1= "SELECT* from testdata ORDER BY 1  DESC ";
                                         $run= mysql_query($query1) ;

                                         while 
                                            ($row =mysql_fetch_array($run)){

                                                $Id=$row['0'];
                                                $name=$row['1'];
                                                $company=$row['2'];
                                                $designation=$row['3'];
                                                $email= $row['4'];
                                                $message=$row['5'];




                                       ?>

<td> <?php echo $Id?>  </td>
<td> <?php echo $name; ?> </td>
<td> <?php echo $company ?> </td>
<td> <?php echo $designation ?> </td>
<td> <?php echo $email ?> </td>
<td> <?php echo $message ?> </td>


<td > <form method ="post" action= "update.php"> <input type ="submit" value= "approve" name ="approve"></input></form>  </td>
</tr>

<?php } ?>
















<tr align='center'>

<td> <?php echo $Id?>  </td>
<td> <?php echo $name1; ?> </td>
<td> <?php echo $company1 ?> </td>
<td> <?php echo $designation1 ?> </td>
<td> <?php echo $email1 ?> </td>
<td> <?php echo $message1 ?> </td>


<td > <input class='action' type ="submit" value= "update" name ="update"></input> </td>
</tr>

Answer 1

只需更改您的查询

"select * from testdata where status =".$_POST['data1'] " 

并在您的脚本中将 $(this).attr('id') 更改为 $(this).val()

  $('#approve').change(function(){
       var data1 = $(this).val();
        $.ajax({ / / ajax block starts
        url: 'update.php', // 
       type: 'POST',
       data: {
         data1: data1 
       },

     success: function(data) {

        }
     });