[error_22]

Hi, I have a problem:

PHP Code:

echo "<form action='send.php' method='POST'>";
echo "<select name='pageid'>";

$sql = "SELECT * FROM `page` ORDER by `id`";
    $result = mysql_query($sql) or die(mysql_error());
    while ($row = mysql_fetch_array($result,MYSQL_ASSOC))
    {
        echo "<option value='{$row['id']}'>{$row['title']}</option>";
    }
    
    echo "</select><input type='submit' value='save'></form>"; 


in send.php:

PHP Code:

$id = $_POST['pageid'];
    $sql = "SELECT * FROM `page` WHERE ?id? = '$id'";
    $result = mysql_query($sql) or die(mysql_error());
    while ($row = mysql_fetch_array($result,MYSQL_ASSOC))
    { echo "something goes here"; } 


I get this when i hit the save button:

You have an error in your SQL syntax. Check the manual that corresponds to your MySQL server version for the right syntax to use near '?id? = '16'' at line 1

What am i doing wrong?