php - SQL 只更新最后一行

Question

我在更新数据库中的多行时遇到问题。这是我的表格。

<form action="results-action" method="post" enctype="multipart/form-data">
<fieldset>   

<table id ="table_id" class="display">

<thead>

<tr><td><h2>Pending Order</h2></td></tr>

<tr>

<th scope="col">Order ID</th>
<th scope="col">Name</th>
<th scope="col">Address</th>
<th scope="col">Product Name</th>
<th scope="col">Produt Quantity</th>
<th scope="col">Price</th>
<th scope="col">Order status</th>
</tr>
</thead>

<tbody>

<?php 

while ($row = mysqli_fetch_array($result)) {
?>

<tr>

<td><input type="text" value='<?=$row['virtuemart_order_id']?>' name="orderid" id="virtuemart_order_id"></td>

<td><?=$row['first_name']?></td>
<td><?=$row['address_1']?></td>
<td><?=$row['order_item_name']?></td>
<td><?=$row['product_quantity']?></td>
<td><?=$row['product_final_price'] ?></td>

<td><select name='change'>

<option value='P'> Pending</option>

<option value='C'> Confirmed</option></select></td>

</tr>

<?php

}

?>

</tbody>
</table>
</fieldset>

<fieldset>
<table>
<tr>
<td><input type="submit" value="Update status" name="update status"> </td>
</tr>
</table>
</fieldset>

</form>

当我更改订单状态以确认每一行时，它应该将每一行的订单状态更新为“P”。但相反，它只更新最新的行。

这是我的更新代码。

$change = $_POST["change"];
$orderid = $_POST['orderid'];

// build SQL statement 
$query = "update ruj3d_virtuemart_order_items set order_status= '$change' WHERE virtuemart_order_id = '$orderid'";

// execute SQL statement
$status = mysqli_query($link, $query) or die(mysqli_error($link));

if ($status) {
echo "<big>It has been updated !!. </big>";

}else {
echo "<br>Has not been successfully confirmed</br>";
}

Answer 1

如下更改您的下拉列表

<select name='change[<?=$row['virtuemart_order_id']?>]'>
  <option value='P'> Pending</option>
  <option value='C'> Confirmed</option>
</select>

然后在你的php中，

foreach($_POST['change'] as $orderid => $change) {
   $query = "update ruj3d_virtuemart_order_items set order_status= '$change' WHERE virtuemart_order_id = '$orderid'";

   // execute SQL statement
   $status = mysqli_query($link, $query) or die(mysqli_error($link));
}