Php drop down menu

i am using following code to insert data into mysql database

<?php
$refselect  = $_POST['refselect'];
$refname    = $_POST['refname'];
$refemail   = $_POST开发者_开发问答['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();
}

?>

"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 accordingly

what to do now ?

You have to fill the dropdown menu on that another page using MySQL select like:

$request = mysql_query("SELECT refselect FROM refdb");

then iterate through all the results to fill your combobox:

echo "<SELECT>";
while ($drow = mysql_fetch_assoc($request))
{
   echo "<OPTION>" . $drow['refselect'] . "</OPTION>";
}
echo "</SELECT>";

Or something like this