Syntax to REPLACE a record in a PHP/MySQL query?

I'm trying to set up a utility where the storeowner can choose from one of the specials she has previously entered into the database through an admin area.

I've got a dropdown menu that is populated with a field from one table. When selected, it should REPLACE the only record in another table. But nothing happens, and I get no warnings or errors.

I can do this through a static dropdown menu, but I want the menu to be generated dynamically.

If I code values in, the table is changed. But I don't seem to be able to figure out how to get my variables into that table.

This is the form:

<form name="form1" id="form1" action="" method="post">

<select name="special" id="special">

<?php

foreach ($conn->query($sql) as $row) {

?>

<option value="<?php echo $row['special']; ?>"><?php echo $row['title']; ?></option>

<?php } ?>

</select>

<input type="submit" name="edit" id="edit" value="Change now">

</form>

and this is the query:

if (array_key_exists('edit', $_POST)) {

// prepare update query

$sql = "REPLACE INTO special VALUES(1, 'placeholder')";

$stmt = $conn->prepare($sql);

// execute query by passing array of variables

$done = $stmt->execute(array($_POST['special']));

}

Where 1 is the record number I want to REPLACE and 'placeholder' represents the variable that will replace what is in the current table, if I knew how to express it properly.

Or should I be doing this a different way? I am using PDO, please no suggestions on mysql or mysqli. I don't know how to convert them and my server does not support them.