php - 如何使用 PHP 和 MYSQL 从表中获取信息到另一个页面？

Question

我在将表格中的信息连接到其他页面时遇到问题。该表由信息组成，右侧有一个名为“查看信息”的按钮。当用户单击该按钮时，它将重定向到 view_order.php。在 view_order.php 中，是表中的信息。

请帮我解决这个问题。 :)

<?php
session_start();
$conn = mysqli_connect('localhost','root','','sampsix');
if(mysqli_connect_errno()){
    echo 'Failed to connect: '.mysqli_connect_error();
}

if(isset($_POST['delete'])){
    $DeleteQuery = "DELETE FROM orders WHERE id='$_POST[hidden]'";
    mysqli_query($conn,$DeleteQuery);
}
if(isset($_POST['view'])){
    header('Location: view_order.php');
}

$query = "SELECT * FROM orders ORDER BY id";
$results = mysqli_query($conn,$query);

echo '<table border="1">';
    echo '<tr>';
        echo '<th>ID</th>';
        echo '<th>Firstame</th>';
        echo '<th>Lastname</th>';
        echo '<th>Email</th>';
        echo '<th>Order Name</th>';
        echo '<th>Order Code</th>';
        echo '<th>Order Qty</th>';
        echo '<th>Sub Total</th>';
    echo '</tr>';

    while($orderData = mysqli_fetch_array($results)){
        echo '<form action="order.php" method="POST">';
        echo '<tr>';
            echo '<td>'.$orderData['id'].'</td>';
            echo '<td>'.$orderData['firstname'].'</td>';
            echo '<td>'.$orderData['lastname'].'</td>';
            echo '<td>'.$orderData['email'].'</td>';
            echo '<td>'.$orderData['ordername'].'</td>';
            echo '<td>'.$orderData['ordercode'].'</td>';
            echo '<td>'.$orderData['orderqty'].'</td>';
            echo '<td>'.$orderData['subtotal'].'</td>';

            echo '<td><input type="hidden" name="hidden" value="'.$orderData['id'].'"></td>';
            echo '<td><input type="submit" name="delete" value="Delete"></td>';
            echo '<td><input type="submit" name="view" value="View Order"></td>';
        echo '</tr>';
        echo '</form>';
    }


echo '</table>';

mysqli_close($conn);

?>

这里是view_order.php的代码

<?php
session_start();
include_once('config.php');

if(isset($_SESSION['id'])){
    $id = $_SESSION['id'];

$query = $mysqli->query("SELECTid,firstname,lastname,email,ordername,ordercode,orderqty,subtotal FROM orders WHERE id='$id'");

if($query){
        while($obj = $query->fetch_object()){
            echo 'id: '.$obj->id;
            echo 'Firstname: '.$obj->firstname;
            echo 'Lastname: '.$obj->lastname;
            echo 'Email: '.$obj->email;
            echo 'Order Name: '.$obj->ordername;
            echo 'Order Code: '.$obj->ordercode;
            echo 'Order Qty: '.$obj->orderqty;
            echo 'Sub Total: '.$obj->subtotal;
        }
    }
}

?>

没有出现在 view_order.php 上。如果我的代码有错误，请检查并提出建议。

Answer 1

我认为不需要 FORM 以防万一。据我了解，您需要在单击“查看”按钮时重定向到详细页面。您可以使用超链接并通过 URL 传递 id，而不是按钮。

  while($orderData = mysqli_fetch_array($results)){
       // echo '<form action="order.php" method="POST">';
        echo '<tr>';
            echo '<td>'.$orderData['id'].'</td>';
            echo '<td>'.$orderData['firstname'].'</td>';
            echo '<td>'.$orderData['lastname'].'</td>';
            echo '<td>'.$orderData['email'].'</td>';
            echo '<td>'.$orderData['ordername'].'</td>';
            echo '<td>'.$orderData['ordercode'].'</td>';
            echo '<td>'.$orderData['orderqty'].'</td>';
            echo '<td>'.$orderData['subtotal'].'</td>';

           // echo '<td><input type="hidden" name="hidden" value="'.$orderData['id'].'"></td>';
            //echo '<td><input type="submit" name="delete" value="Delete"></td>';
            echo "<td><a href='view_order.php?id=".$orderData['id']."'>View</a></td>";
        echo '</tr>';
        //echo '</form>';
    }

并在 view_order.php 中

<?php
include_once('config.php');

if(isset($_GET['id'])){
    $id = $_GET['id'];

$query = $mysqli->query("SELECT id,firstname,lastname,email,ordername,ordercode,orderqty,subtotal FROM orders WHERE id='$id'");

if($query){
        while($obj = $query->fetch_object()){
            echo 'Firstname: '.$obj->firstname;
        }
    }
}

?>