Drop down no result? Java+php

mixed java php code

[Reverentx]

here is my drop down (work fine)
<?php
if(isset($_POST['submit'])) {
} else {
echo'<SELECT ID="type" NAME="type">';
echo'<OPTION VALUE="">-----CHOOSE-----</OPTION>';
include'open.php';
$sql= "SELECT * FROM contractortype ORDER BY cono DESC";
$result= mysql_query($sql,$db);
$row= mysql_fetch_array($result);
$cono=$row["cono"];
$value='1';
while ($cono>'0') {
$sql= "SELECT cono,type FROM contractortype WHERE cono='$cono'";
$result= mysql_query($sql,$db);
$row= mysql_fetch_array($result);
$type=$row["type"];
if ($type) {
echo'<OPTION VALUE"'.$value.'">'.$type.'</OPTION>';
$value++;
}
$cono=$cono-100;
}
echo'</SELECT>';
}
?>
on the same page i Refresh the page to get the result
it dont work? what wrong
i have worked with it for 3 dayes now without result?
HELP? some one have a simple solution!
<?php
if(isset($_POST['submit'])) {
$type=($_POST['type']);
if (isset($_POST['type'])) { $ok="EXIST"; }
echo "value: ", $opt," Type:",$type, " ok: ",$ok; }
?>
<form method="post"><input type="submit" onClick=popup() name="submit" value="submit" /></form>

[StupidScript]

Please let us know (a) what you expect from this code and (b) what is "not working" about it. There are some issues with the way it's written, but it's a little hard to tell what you are trying to do. It looks like you want to populate a SELECT form element with items from a database ...?

Also, where is the "popup()" function being defined? It's the only Javascript in the code you posted.

[coopster]

Welcome to WebmasterWorld, Reverentx!

Also, you don't have your <select> inside your <form> tags.