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\"> PHP: <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> PHP:
Use ajax - it's easy with jquery and prototype. decide which one you want to use and then look at their examples.
I am not quite sure what you want but what I understand is you want insert area_names into select and when the select change this javascript code runs: jwplayer().load({file:'http://www.onfilm.biz/streaming/home/area_films/ ".$area_name.".flv'});jwplayer().play(true); <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"); $Echos = " <select name='aid' size='1' onchange='jwplayer().load({file:\"http://www.onfilm.biz/streaming/home/area_films/\"+this.value+\".flv\"});jwplayer().play(true);' style='width:160px'> <option value='none'>...location films...</option>"; while ($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)"; } $Echos .= "<option value='{$area_name}'>{$area_name}</option>"; } echo $Echos; mysql_close($connectionat); ?> </select> <input style="width:70px;" type="submit" name="view" value="SEARCH" /> </form> PHP: