Question

<?php
$sample = 0;
    if($sample ==0)
    {   
        $displayinner = '<option value =' . $row['privilege_ID'] . '>' . $row['privilege_name'] . '</option>';
        $displayouter = '<td><label>Privileges:</label></td>
          <td><select name = "Privilege" id = "Privilege" multiple="multiple">
              <?php
                $PrivilegeNames = mysql_query("SELECT * FROM privilege");
                while($row = mysql_fetch_array($PrivilegeNames))
                    echo $displayinner;
              ?>
            </select></td>
          </tr>
          <tr>';
        echo $displayouter;
    }
?>

I only get a dropdownbox without any data inside it. Please help, what am i doing wrong?

Answer 1

Yeah that won’t work – there’s a lot of issues, your “template” code will be parsed which will cause errors, and you can’t nest <?php blocks like that. To get what you want, try this for starters:

$displayouter = '<td><label>Privileges:</label></td><td><select name = "Privilege" id = "Privilege" multiple="multiple">';

$PrivilegeNames = mysql_query("SELECT * FROM privilege");
while($row = mysql_fetch_array($PrivilegeNames)) {
    $displayouter .= '<option value =' . $row['privilege_ID'] . '>' . $row['privilege_name'] . '</option>';
}

$displayouter .= '</select></td></tr><tr>';

echo $displayouter;

If you don’t need the variable $displayouter later, you can just echo everything instead, or break out of the PHP block entirely if this is template code. Note the use of .=, which appends data to the existing variable.

There’s a lot of possible security holes here, but hopefully this helps to see some working code. Make sure to use htmlspecialchars on all unknown HTML output, for one thing.