Question

我有一个使用用户输入的表单，并与数据库中的相同值进行比较，如果输入无效，则拒绝并且不更新。

当用户输入正确时，表单会转发它，我的代码会从该输入中找到相应的ID，然后更新它。

但到目前为止，它的作用是更新最后一条记录。

这是我的代码：

我道歉它有点长，所以请耐心等待。我仍然是学生和总菜鸟。

require("connect.php");
$query = mysql_query("SELECT * FROM driver LEFT JOIN vehicle ON driver.vehicle_id = vehicle.vehicle_id");
while($driverlist = mysql_fetch_array($query)){

                            $vehiclestatus = $driverlist['v_status'];
                            $driver = $driverlist['d_fname'];
                            $driver_id =$driverlist['driver_id'];
                            $v_model = $driverlist['v_model'];
                            $plate_number = $driverlist['plate_number'];
                            $vehicleid = $driverlist['vehicle_id'];
                            echo '<tr>';
                            echo '<td>'.$driver.'</td>';

                            echo '<td>'.$driver_id.'</td>';

                            echo '<td>'.$v_model.'</td>';
                            echo '<td>'.$plate_number.'</td>';







                        echo '</tr>';
}









if(isset($_POST['edit'])){
    $vehicle_id = $_POST['vehicle_id'];
    $driverid = $_POST['driver_id'];

               require("connect.php");
                       mysql_query("UPDATE vehicle RIGHT JOIN driver ON vehicle.vehicle_id = driver.driver_id  SET v_status = '1'   WHERE driver.vehicle_id = '".$vehicle_id."' AND driver_id = '".$driverid."'")
                or die(mysql_error());

                   }

if(isset($_POST['update']))
                {
                   $vehicle_id = $_POST['vehicle_id'];
                   $driverid = $_POST['driver_id'];

                       require("connect.php");

                mysql_query("UPDATE vehicle RIGHT JOIN driver ON vehicle.vehicle_id = driver.driver_id SET v_status = '0' WHERE driver.vehicle_id = '".$vehicle_id."' AND driver_id = '".$driverid."'")
                or die(mysql_error());




                header('Location: drivers.php');
                }


if(!empty($_POST['registerbtn']))
{
    $driver = $_POST['driver'];
    require("connect.php");


                        $match_driver = mysql_query("SELECT * FROM driver WHERE d_fname = '$driver'");
                        $rows = mysql_num_rows($match_driver);






                        if($rows != 0)
                        {
                            while($row_driver = mysql_fetch_array($match_driver))
                            {


                                $a_fname = $row_driver[2];
                                $a_id = $row_driver[0];
                            }
                        }
                        error_reporting(0);
                        if($driver == $a_fname){

                        require("connect.php");

                        mysql_query("UPDATE trip_reservation SET driver_id = '".$a_id."' WHERE trip_id = '".$tripid."'");

                        //mysql_query("INSERT INTO trip_reservation (trip_id, driver_id) VALUES('".$tripid."', '".$a_id."') ON DUPLICATE KEY UPDATE trip_id = '".$tripid."'")or die(mysql_error());



                        echo "Driver Added Successfully";

                        }
                        else{
                            echo "The Driver is not found in the database <br>
                            Please input a valid Driver.";
                        }


}




else {
    echo $registerform;
    }
?>

Answer 1

为什么在更新单个表上的记录时连接表。其次，你在比较你甚至没有更新的表格上的WHERE条款

而不是这个

"UPDATE vehicle RIGHT JOIN driver ON vehicle.vehicle_id = driver.driver_id SET v_status = '0' WHERE driver.vehicle_id = '".$vehicle_id."' AND driver_id = '".$driverid."'"

这样做

"UPDATE vehicle SET v_status = '0' WHERE id = '".$vehicle_id."' AND driver_id = '".$driverid."'"

这假设您的车辆表有一个名为 id 的主键字段和一个名为 driver_id

的drivers_table的外键字段

是的，它应该更新JUST ONE RECORD

PHP Update仅更新最后一条记录

1 个答案: