dropdown list with ajax

ajax drop down menu is easy to learn,
i just find how to create dropdown list.

here is the example

For This Example, I using PHP Code and MySQL Database
1. Create Dropdown List (state_dropdown.php)

<form name=sel>
States: <font id=states><select>
<option value=""> ----- </option>
</select></font>
     
Cities : <font id=cities><select>
<option value='0'<=== none ===>/option>
</select$gt;</font>
?$gt;

<script language=Javascript>
function Inint_AJAX() {
try { return new ActiveXObject("Msxml2.XMLHTTP");  } catch(e) {} //IE
try { return new ActiveXObject("Microsoft.XMLHTTP"); } catch(e) {} //IE
try { return new XMLHttpRequest();          } catch(e) {} //Native Javascript
alert("XMLHttpRequest not supported");
return null;
};

function dochange(src, val) {
var req = Inint_AJAX();
req.onreadystatechange = function () {
 if (req.readyState==4) {
      if (req.status==200) {
           document.getElementById(src).innerHTML=req.responseText; //retuen value
      }
 }
};
req.open("GET", "state.php?data="+src+"&val="+val); //make connection
req.setRequestHeader("Content-Type", "application/x-www-form-urlencoded;charset=iso-8859-1"); // set Header
req.send(null); //send value
}

window.onLoad=dochange('states', -1);         // value in first dropdown
</script>

2. Select States and Cities to Show in Dropdown (state.php)

<?
$data=$_GET['data'];
$val=$_GET['val'];
//set database
$dbhost = "localhost";
$dbuser = "";
$dbpass = "";
$dbname = "test";
mysql_pconnect($dbhost,$dbuser,$dbpass) or die ("Unable to connect to MySQL server");
if ($data=='states') {  // first dropdown
  echo "<select name='states' onChange=\"dochange('cities', this.value)\">";
  echo "<option value='0'>==== choose state ====</option>";
  $result=mysql_db_query($dbname,"select `id`, `state` from states order by `state`");
  while(list($id, $name)=mysql_fetch_array($result)){
       echo "<option value=\"$id\" >$name</option> \n" ;
  }
} else if ($data=='cities') { // second dropdown
  echo "<select name='cities' >\n";
  echo "<option value='0'>====choose cities ====</option>\n";
  $result=mysql_db_query($dbname,"SELECT `id`, `city` FROM cities WHERE `state_id` = '$val' ORDER BY `city` ");
  while(list($id, $name)=mysql_fetch_array($result)){
       echo "<option value=\"$id\" >$name</option> \n" ;
  }
}
echo "</select>\n";
?>

Preview: