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>