Update multiple rows without changing id value - php

Question

We have Table like below :
Once we click on "Submit" button , I am displaying random numbers below column "tracking id" using below query , its working fine:
$sql = $con->query("update orders set tracking_id = '$r' WHERE id ='1'");
$sql = $con->query("update orders set tracking_id = '$r' WHERE id ='2'");
But when i use below query, its not updating....
$sql = $con->query("update orders set tracking_id = '$r' WHERE id ='$id'");
Code :
<?php
$result = mysqli_query($con,"SELECT * FROM orders");
echo "<table border='1'>
<tr>
<th>order</th>
<th>payment</th>
<th>generate</th>
<th>tracking id</th>
</tr>";
while($row = mysqli_fetch_array($result))
{
$id = $row['id'];
echo "<tr>";
echo "<td>" . $row['order_id'] . "</td>";
echo "<td>" . $row['payment_type'] . "</td>";
echo "<td><form method='post' action='new1.php'>
<input type = submit>
</form> </td>";
echo "<td>" . $row['tracking_id'] . "</td>";
echo "</tr>";
}
echo "</table>";
?>
new1.php
<?php
$id = $row['id'];
$r = mt_rand(1000,9999);
$sql = $con->query("update orders set tracking_id = '$r' WHERE id ='$id'");
?>

Answer 1

I think that you're not including the details in the form
<?php
$result = mysqli_query($con, "SELECT * FROM orders");
echo "<table border='1'>
<tr>
<th>order</th>
<th>payment</th>
<th>generate</th>
<th>tracking id</th>
</tr>";
while ($row = mysqli_fetch_array($result)) {
$id = $row['id'];
echo "<tr>";
echo "<td>" . $row['order_id'] . "</td>";
echo "<td>" . $row['payment_type'] . "</td>";
echo "<td>";
if (empty($row['tracking_id'])) {
echo "<form method='post' action='new1.php'>";
echo "<input type ='hidden' name='id' value='$id'>
<input type='submit'>
</form>";
}
echo "</td>";
echo "<td>" . $row['tracking_id'] . " </td > ";
echo "</tr>";
}
echo "</table >";
?>
And you also need to modify new1.php
<?php
$id = $_POST['id'];
$r = mt_rand(1000,9999);
$sql = $con->query("update orders set tracking_id = '$r' WHERE id ='$id'");
?>