How can i insert drop down option values into the select

Hi i have this form which the drop down options value is based on database query, i want to when onchange to add the drop down option value into the select value but i cant seem to be able to grab the drop down values, the drop down variable is $area_name i want to add from the drop down option to
here

echo " <select name=\\"aid\\" size=\\"1\\" onchange=\\"jwplayer().load({file:'http://www.onfilm.biz/streaming/home/area_films/ ".$area_name.".flv'});jwplayer().play(true);\\" style=\\"width:160px\\"> 
        <form id="area_films" action="<?php echo $_SERVER['PHP_SELF']; ?>"  method="post" name="area_films">
                    <?php
                    //get the DB connection variables
                    include("includes/config.php");
                    //connect to DB
                    $connectionat = @mysql_connect($db_address,$db_username,$db_password) or die("Couldn't CONNECT.");
                    $dbcat = @mysql_select_db($db_name, $connectionat) or die("Couldn't select DATABASE.");
                    //SELECT or FIND the same PASSWORD
                    $queryat="SELECT * FROM films_area WHERE (area_type = 'town' OR area_type = 'village' OR area_type = 'city') ORDER BY area_town ASC";
                    $resultat = mysql_query($queryat) or die("Couldn't execute QUERY - FIND Client TVs");
					
echo " <select name=\\"aid\\" size=\\"1\\" onchange=\\"jwplayer().load({file:'http://www.onfilm.biz/streaming/home/area_films/ ".$area_name.".flv'});jwplayer().play(true);\\" style=\\"width:160px\\">
                <option value=\\"none\\">...location films...</option>";
                        while ($rowat = mysql_fetch_array($resultat))
                            {
                                $area_id=$rowat['area_id'];
                                $area_name=$rowat['area_town'];
                                if($rowat['area_country'] == 'France')
                            {
                                 $area_name =  $area_name."(F)";
                            }
                                print("<option value='$area_id'>$area_name</option>");
                            }
                                mysql_close($connectionat);
                        ?>
            </select>
                <input style="width:70px;" type="submit" name="view" value="SEARCH" />
        </form>

So if I understand correctly, you want to change the value in the <select> tag when the user chooses another option from the drop down?
You can’t do that with PHP. You’ll have to use Javascript to get the selected value, and put it where you want to use it.

It looks like you want to play the selected video file (.flv).

If that is the case, just call a function in the <select>'s onchange event handler to which you pass the selected option’s value (this.value). The event handler function then runs the code to play the selected video.