Multiple Selection Menu Dilemma

Okay I have the following code so far.

        $sql = "SELECT stass_warrior FROM staff_assignments WHERE stass_pos = $pos_id AND stass_weekend = $initiation";
        if (!$query = mysql_query($sql))        error(mysql_error());
        $query = mysql_fetch_row($query);

        $sql = "SELECT war_id, CONCAT(war_lname, ', ', war_fname) FROM warrior, staff_roster WHERE war_id != 1 AND staff_weekend = $initiation AND staff_warrior=war_id ORDER BY war_lname";
        if (!$res = mysql_query($sql))  error(mysql_error());

        tr();
        echo TTT . '<th class="format" width="25%">' . $row[1] . '</th>' . B;
        echo TTT . '<td class="format">' . B;
        echo TTT . '<select name="position_' . $n . '[]" size="5" multiple>' . B;
        while($rec = mysql_fetch_row($res))     {
                ($query[0] == $rec[0] ? $selected =" selected" : $selected = "");
                printf("\t\t\t\t<option value=\"%s_%s\"%s>%s</option>\n", $pos_id, $rec[0], $selected, $rec[1]);
        }
        echo TTT . '</select>' . B;
        echo TTT . '</td>' . B;
        trc();

Now my problem is with the multiple feature of the drop down menu. The first query to the db resulting in the $query variable may contain more then one result. However only the last result will appear with the selected option. I need some code that is is going to display the selected option next to all the correct results in the drop down. Understand what I am needing? Thanks for your help Phillip