php - 如何更新下拉列表中的特定行？

Question

我想在选择下拉列表中的选项后更新表格。但是，当我点击批准按钮时，数据库中的车牌没有更新。有人能帮我吗？谢谢。

<?php 

    mysql_connect("l","t","")or die(mysql_error()); // connect to server
    mysql_select_db("car")or die("Cannot connect to database") ;
    $query=mysql_query("Select * from list "); //SQL Query

                echo "<table><tr>
                    <th>Name</th>
                    <th>Destination</th>
                    <th>Time</th>
                    <th>Date</th>
                    <th>Car Requested</th>
                    <th>Purpose</th>
                    <th>Car Approve</th>
                    <th>Approve</th>
                    <th>Reject</th>
                </tr>"; 


    while($row=mysql_fetch_array($query))
    {
        if($row['status'] =='Pending'){

            echo"<tr>";
            echo'<td align="center">'.$row['employee']."</td>";
            echo'<td align="center">'.$row['destination']."</td>";
            echo'<td align="center">'.$row['time']."</td>";
            echo'<td align="center">'.$row['date']."</td>";
            echo'<td align="center">'.$row['car']."</td>";
            echo'<td align="center">'.$row['message']."</td>";

            echo'
             <form action="approve.php" method="GET">
                <td align="center">
                <select name="car_plate">
                <option>--Select Car--</option> 
                <option>WBN 3041</option>
                <option>WWW 4545</option>
                <option>BBB 1111</option>
                <option>CCC 2222</option>
                <option>DDD 3333</option>
                </select>
                 </td>
            </form>
            ';

            echo'<td align="center"><a href="#" onclick="fapprove('.$row['id'].')">Approve</a> </td>';  
            echo'<td align="center"><a href="#" onclick="freject('.$row['id'].')">Reject</a> </td>';    
            echo"<tr>";
        }

    }
echo"</table>"; 

?>
<script>
            function fapprove(id)
            {
                var r=confirm("Are you sure you want to APPROVE this record?");
                if(r==true)
                {
                    window.location.assign("approve.php?id=" + id);
                    alert('APRROVED! =)');
            }

            }
        </script>

批准.php

    if($_SERVER['REQUEST_METHOD'] == "GET")
{
    mysql_connect("l", "r","") or die(mysql_error()); //Connect to server
    mysql_select_db("car") or die("Cannot connect to database"); //Connect to database
    $id = $_GET['id'];
    $car_plate = $_GET['car_plate'];

    mysql_query("UPDATE list SET status='Approved' WHERE id='$id'");
    mysql_query("UPDATE list SET car_plate='$car_plate' WHERE id='$id'");

Answer 1

您没有在 javascript 函数中传递车牌号。使用下面的代码

     function fapprove(id)
        {
            var r=confirm("Are you sure you want to APPROVE this record?");
            if(r==true)
            {
                var carPlate = $("#car_plate").val();
                window.location.assign("approve.php?id="+id+"&car_plate="+carPlate);
                alert('APRROVED! =)');
        }

        }

将 id 添加到选择下拉列表中作为 <select id="car_plate" name="">

希望它对你有用。如果您需要任何帮助，我随时准备指导您...