如何使用PHP和MYSQL從表中獲取信息到另一個頁面？

Question

我有問題將表中的信息連接到其他頁面。該表由信息組成，並在名爲「查看信息」的右側有一個按鈕。當用戶點擊該按鈕時，它將重定向到view_order.php。裏面的view_order.php，是表中的信息。

請幫我弄清楚這一點。 :)

<?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); 

?> 

這裏是view_order.php

<?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; 
     } 
    } 
} 

?> 

無的代碼出現在view_order.php。如果我的代碼有錯誤，請檢查它並建議。

Answer 1

我不認爲FORM是需要的情況下。據我瞭解，當你點擊查看按鈕時，你需要重定向到詳細頁面。您可以使用超鏈接並通過URL傳遞該ID，而不是按鈕。

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>'; 
    } 

和view_order.php

<?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; 
     } 
    } 
} 

?>