PHP 不会通过 POST 传递变量

Question

I am having a issue where my PHP code isn't passing the variable in the input. The value outputs as just a string of "<?php echo $orderID?>" . What I would like it to do actually send the variable created.

<?php 
if ($order['status'] != "Received") {
    $orderID = $order['orderID'];
    var_dump($orderID);
    echo '<td><form action="./markAsReceived.php" method="POST">
          <input type="hidden" name="orderID" value="<?php echo $orderID; ?>">
          <input type="submit" class="btn btn-secondary mx-1" value="Mark As Received">
          </form></td>';
} else {
    echo '<td></td>';
}
?>

So from this the code checks the database if an Orders status does not equal Received then it displays a button so you can mark it received otherwise it displays nothing. The button does appear but like I said before the value for the button is just the string of the php code.

I'm sure it's something simple and I just can't wrap my mind around it, thank you in advance for the help!

Answer 1

You are not using correct string concatenation. In double quotes, php looks for variables, but in single quotes it doesn't. As you are using html that uses double quotes, single quotes have to be used for your string. You can concatenate within a string:

<?php 
if ($order['status'] != "Received") {
$orderID = $order['orderID'];
var_dump($orderID);
echo '<td><form action="./markAsReceived.php" method="POST">
      <input type="hidden" name="orderID" value="'.$orderID.'">
      <input type="submit" class="btn btn-secondary mx-1" value="Mark As Received">
      </form></td>';
} else {
echo '<td></td>';
}
?>