i am using following code to insert data into mysql database\[code\]<?php$refselect = $_POST['refselect'];$refname = $_POST['refname'];$refemail = $_POST['refemail'];$refcnt = $_POST['refcnt'];$refdes = $_POST['refdes'];$referror = $cberror = "<h1>Data not Added</h1><br/><br/><h3>Please Follow The Instructions</h3>";$urefdb = "INSERT INTO refdb(reftype,refname,refemail,refcnt,refdes) VALUES ('$refselect','$refname','$refemail','$refcnt','$refdes');"; include_once("db.php");if ($refselect == "Select Type") die ("$referror");if (empty ($refname)) die ("$referror");if (mysql_query("$urefdb")){echo "<h3>One Record Updated Successfully with the following Details </h3><br/>";echo "Reference Type =$refselect <br/><br/>";echo "Reference Name = $refname <br/><br/>";echo "Reference E-Mail = $refemail <br/><br/>";echo "Reference Description = $refdes <br/><br/>";}else{echo mysql_error();}?>\[/code\]"refselect" data is coming from a drop down menu at the page now i want that as i add data through this form another form located at another page pic "refname" from database as i update "refdb" that drop down menu pick data accordinglywhat to do now ?