Help me create AJAX - Dropdown?

I create dropdow list like:

1)db_ajax.sql

# phpMyAdmin MySQL-Dump
# version 2.3.2
# http://www.phpmyadmin.net/ (download page)
#
# Host: localhost
# Generation Time: Aug 14, 2006 at 09:14 PM
# Server version: 4.00.00
# PHP Version: 4.2.3
# Database : `db_ajax`
# --------------------------------------------------------

#
# Table structure for table `city`
#

CREATE TABLE city (
  id tinyint(4) NOT NULL auto_increment,
  city varchar(50) default NULL,
  countryid tinyint(4) default NULL,
  PRIMARY KEY  (id)
) TYPE=MyISAM;

#
# Dumping data for table `city`
#

INSERT INTO city VALUES (1, 'Los Angales', 1);
INSERT INTO city VALUES (2, 'New York', 1);
INSERT INTO city VALUES (3, 'Toranto', 2);
INSERT INTO city VALUES (4, 'Vancovour', 2);

# --------------------------------------------------------

#
# Table structure for table `country`
#

CREATE TABLE country (
  id tinyint(4) NOT NULL auto_increment,
  country varchar(20) NOT NULL default '',
  PRIMARY KEY  (id)
) TYPE=MyISAM;

#
# Dumping data for table `country`
#

INSERT INTO country VALUES (1, 'USA');
INSERT INTO country VALUES (2, 'Canada');

[B]2) findcity.php[/B]
<?
#### Roshan's Ajax dropdown code with php
#### Copyright reserved to Roshan Bhattarai - [email]nepaliboy007@yahoo.com[/email]
#### if you have any problem contact me at http://roshanbh.com.np
#### fell free to visit my blog http://php-ajax-guru.blogspot.com
?>

<? $country=$_REQUEST['country'];
$link = mysql_connect('localhost', 'root', ''); //changet the configuration in required
if (!$link) {
    die('Could not connect: ' . mysql_error());
}
mysql_select_db('db_ajax');
$query="select city from city where countryid=$country";
$result=mysql_query($query);

?>
<select name="city">
<option>Select City</option>
<? while($row=mysql_fetch_array($result)) { ?>
<option value><?=$row['city']?></option>
<? } ?>
</select>

[B]3) index.php[/B]
<html>
<head>
<title>Roshan's Ajax dropdown code</title>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1">
<?
#### Roshan's Ajax dropdown code with php
#### Copyright reserved to Roshan Bhattarai - [email]nepaliboy007@yahoo.com[/email]
#### if you have any problem contact me at http://roshanbh.com.np
#### fell free to visit my blog http://php-ajax-guru.blogspot.com
?>
<script>
function getXMLHTTP() { //fuction to return the xml http object
		var xmlhttp=false;	
		try{
			xmlhttp=new XMLHttpRequest();
		}
		catch(e)	{		
			try{			
				xmlhttp= new ActiveXObject("Microsoft.XMLHTTP");
			}
			catch(e){
				try{
				xmlhttp = new ActiveXObject("Msxml2.XMLHTTP");
				}
				catch(e1){
					xmlhttp=false;
				}
			}
		}
		 	
		return xmlhttp;
	}
	
	
	
	function getCity(strURL) {		
		
		var req = getXMLHTTP();
		
		if (req) {
			
			req.onreadystatechange = function() {
				if (req.readyState == 4) {
					// only if "OK"
					if (req.status == 200) {						
						document.getElementById('citydiv').innerHTML=req.responseText;						
					} else {
						alert("There was a problem while using XMLHTTP:\
" + req.statusText);
					}
				}				
			}			
			req.open("GET", strURL, true);
			req.send(null);
		}
				
	}
</script>




</head>

<body>

<form method="post" action="" name="form1">
<table width="60%" border="0" cellspacing="0" cellpadding="0">
  <tr>
    <td width="150">Country</td>
    <td  width="150"><select name="country" onChange="getCity('findcity.php?country='+this.value)">
	<option value="">Select Country</option>
	<option value="1">USA</option>
	<option value="2">Canada</option>
        </select></td>
  </tr>
  <tr style="">
    <td>City</td>
    <td ><div id="citydiv"><select name="city">
	<option>Select City</option>
        </select></div></td>
  </tr>
  <tr>
    <td> </td>
    <td> </td>
  </tr>
  <tr>
    <td> </td>
    <td> </td>
  </tr>
</table>
</form>
</body>
</html>
==========

But when I sellect Country , city don’t sellect automatic
HELP ME

Right off the bat, if you are having issues, you should ask the content author as stated in the code you posted:


#### Roshan's Ajax dropdown code with php
#### Copyright reserved to Roshan Bhattarai - nepaliboy007@yahoo.com
#### if you have any problem contact me at http://roshanbh.com.np
#### fell free to visit my blog http://php-ajax-guru.blogspot.com

I don’t like that example so I wrote my own version of it:

<?php

	/* SQL
	
		CREATE TABLE `city` (
		  `id` tinyint(4) NOT NULL AUTO_INCREMENT,
		  `city` varchar(50) DEFAULT NULL,
		  `countryid` tinyint(4) DEFAULT NULL,
		  PRIMARY KEY (`id`)
		) ENGINE=MyISAM;
		
		INSERT INTO `city` VALUES(1, 'Los Angeles', 1);
		INSERT INTO `city` VALUES(2, 'New York', 1);
		INSERT INTO `city` VALUES(3, 'Toronto', 2);
		INSERT INTO `city` VALUES(4, 'Vancouver', 2);
		
		CREATE TABLE `country` (
		  `id` tinyint(4) NOT NULL AUTO_INCREMENT,
		  `country` varchar(20) NOT NULL DEFAULT '',
		  PRIMARY KEY (`id`)
		) ENGINE=MyISAM;
		
		
		INSERT INTO `country` VALUES(1, 'USA');
		INSERT INTO `country` VALUES(2, 'Canada');
	
	*/

	error_reporting(-1);
	
	try {
		$dbh = new PDO(
			'mysql:dbname=test;host=localhost',
			'root',
			'root'
		);
	} catch (PDOException $e) {
		echo 'Connection failed: ' . $e->getMessage();
	}	
	
	function getCountries($db) {
		$list = '';
		$sql = 'SELECT id, country FROM country';
		
		$stmt = $db->query($sql);
		while( $row = $stmt->fetch(PDO::FETCH_ASSOC) ) {
			$list[] = $row;
		}		
		return $list;
	}
	
	function getCitiesFromID($id, $db) {
		$list = '';
		$sql = 'SELECT id, city FROM city WHERE countryid = ?';
		
		$stmt = $db->prepare($sql);
		$stmt->bindParam(1, $id, PDO::PARAM_INT);
		$stmt->execute();
		while( $row = $stmt->fetch(PDO::FETCH_ASSOC) ) {
			$list[] = $row;
		}		
		return $list;		
	}
		
	if( isset($_GET['country']) ) {
		echo json_encode( getCitiesFromID( $_GET['country'], $dbh ) );
		die();
	}

?>
<html>
	<body>
		<form id="testForm">
		
			<label>Country</label>
			<select name="country">
				<option></option>
				<?php
					foreach( getCountries($dbh) as $row ) {
						echo '<option value="' . $row['id'] .'">' . $row['country'] .'</option>';				
					}
				?>
			</select>	
		
			<label>City</label>
			<select name="city"></select>		
		
		</form>
		
		<script src="https://ajax.googleapis.com/ajax/libs/jquery/1/jquery.min.js"></script>
		<script>
			$('#testForm select[name="country"]').on(
				'change keyup',
				function() {
					var $citySelect = $('#testForm select[name="city"]');
						$citySelect.empty();				
					if( parseInt(this.value) ) { // so we don't parse the empty option...
						$.ajax({
							url: 'test.php?country=' + this.value,
							dataType: 'json',
							success: function(data) {
								$.each(data, function(k,v) {
									$citySelect.append('<option value="' + data[k].id + '">' + data[k].city + '</option>');
								});
							}
						});
					}
				}				
			);
		</script>
	</body>
</html>

Thanks you very much!