Php 如何验证用户从mysql数据库中删除记录

我目前有一个预订系统，允许用户查看、编辑和删除记录。系统目前允许用户随时删除记录。但是，由于它是一个预订系统，因此用户只能在预订日期前48小时删除记录，并应限制用户在预订日期后24小时内删除记录。我是否需要包含某种函数

这是我删除记录的当前代码：

bookings.php

   echo "<td><a href = 'delete.php?id=".$row['booking_id']."' onclick='return          

   confirmation()'> Remove </td>"; 


      <script type="text/javascript">

        function confirmation()

  {

        if (confirm("Are you sure you want to cancel booking")){
             location.href='delete.php';

           }

           else {

            return false;


           }
         } 

---

我不知道您的表结构，但它应该是这样的：

DELETE from bookings1 where booking_id='$id' AND booking_id IN (SELECT booking_id FROM bookings1 
WHERE BookingDate >= DATEADD(HOUR, -48, GETDATE())
  AND BookingDate <  DATEADD(HOUR, -24, GETDATE())

也改变

location.href='delete.php'

到

location.href='delete.php？id=.$row['booking_id']；返回false

---

否则$\u GET为空

您可以将日期列添加到数据库中的记录中，然后在删除之前检查还有多少时间可以删除

$yourEntry = mysql_fetch_array(mysql_query("select * from bookings1 where booking_id='$id'"));


//Checking difference between two dates (current and your entry date)       
$diff = strtotime(date('Y-m-d H:i:s')) - strtotime(''.$yourEntry[day].' '.$yourEntry{hour].'');

//this is for less than 30 seconds, change for your value
                if($diff < 30){
                    $problem = TRUE;
                }



        if (!$problem){

//your code where there is not a problem with time. In my code if more than 30 seconds. 

         }

---

我的表结构是booking_id，BookingDate，TimeslotI我尝试过这种方法，但它不会从数据库中删除任何内容我的表由以下字段组成，即booking_id，BookingDate，timeslot@RohanRehman你的预订日期怎么样？它的日期是带时间的还是只带日期的？所以再添加一个带时间的字段。将日期设置为Y-m-d，时间设置为H:i:s。我的代码应该可以处理这个字段。如果我要删除statement@RohanRehman把它放在if$问题{//这里是您的代码}语句

$yourEntry = mysql_fetch_array(mysql_query("select * from bookings1 where booking_id='$id'"));


//Checking difference between two dates (current and your entry date)       
$diff = strtotime(date('Y-m-d H:i:s')) - strtotime(''.$yourEntry[day].' '.$yourEntry{hour].'');

//this is for less than 30 seconds, change for your value
                if($diff < 30){
                    $problem = TRUE;
                }



        if (!$problem){

//your code where there is not a problem with time. In my code if more than 30 seconds. 

         }