javascript - PHP 文件无法从 Jquery ajax 读取多个参数

Question

我已经阅读了该问题的多种解决方案，但似乎没有一个适合我。我有两个文件 manageadmin.php 和rejectaction.php 。 Manageadmin.php 文件为用户提供了一个拒绝管理员的按钮。当某人单击拒绝按钮时，管理员状态必须设置为零。

这是我的初始代码

manageadmin.php

<script type="text/javascript">
$(document).ready(function()
    {
       /* reject */
       $(".reject_btn").click(function(){

         var element = $(this);
         var reject_id = element.attr('reject_id');
         $.ajax({
            type: "POST",
            url: "rejectaction.php",
            data: {reject_id:reject_id},
            success: function(){
                alert("Reject Successful");
                location.reload();
            }
         });

       });
    });
</script>

拒绝操作.php

    $reject_id=$_POST['reject_id'];
    $reject_query="UPDATE tbl_admin set admin_status=0 where admin_id='$reject_id'";
    mysql_query($reject_query);

这工作正常，但后来我意识到我需要多个页面的拒绝代码。所以我想到将where条件中的表名、要更新的列和要检查的列作为参数传递。这是修改后的代码:

<script type="text/javascript">
$(document).ready(function()
    {
       /* reject */
       $(".reject_btn").click(function(){

         var element = $(this);
         var reject_id = element.attr('reject_id');
         var tbl_name = "tbl_admin";
         var column_reject = "admin_status";
         var column_cond = "admin_id";
         $.ajax({
            type: "POST",
            url: "rejectaction.php",
            data: {reject_id:reject_id, tbl_name:tbl_name, 
                column_cond:column_cond,
                column_reject:column_reject},
            success: function(){
                alert("Reject");
                location.reload();
            }
         });

       });
    });
</script>

拒绝操作.php

    $reject_id=$_POST['reject_id'];

    $tbl_name = $_POST['tbl_name'];
    $column_cond = $_POST['column_cond'];
    $column_reject = $_POST['column_reject'];

    $reject_query="UPDATE '$tbl_name' set '$column_reject' = 0 where '$column_cond'='$reject_id'";


// $reject_query="UPDATE tbl_admin set admin_status=0 where admin_id='$reject_id'";
    mysql_query($reject_query);

此代码不起作用(即状态未更新)。但是 alert("Reject Success") 正在执行。我是初学者。有人能指出我所犯的错误吗？谢谢

Answer 1

除了您在评论中收到的(好的)建议之外[另请参阅此答案]，您的主要问题可能是您包装了 $table_name, $column_reject 和引号之间的 $column_cond。

你应该写:

$reject_query =
    "UPDATE $tbl_name set $column_reject = 0 where $column_cond='$reject_id'";