php+jQuery实现的四级联动
前台:
<html>
<head>
	<meta http-equiv="Content-Type" content="text/html;charset=utf-8" />
	<script type="text/javascript" src="jquery-1.7.1.min.js" ></script>
	<script type="text/javascript" >
		$.post("bak.php", {id:0},
			function(status){
				str = ‘<option>请选择</option>’;
				for(var i=0;i<status.length;i++){
					str += ‘<option value="’+status[i].id+’" >’+status[i].name+'</option>’;
				}
			$(‘select’).html(str);
			}, "json");
		function change(obj){
			$(obj).nextAll().remove();
			$.ajax({
				‘url’:’bak.php’,
				‘type’:’post’,
				‘data’:{‘id’:$(obj).val()},
				‘dataType’:’json’,
				‘success’:function(status){
					str = ‘<select onchange="change(this)" ><option>请选择</option>’;
					for(var i=0;i<status.length;i++){
						str += ‘<option value="’+status[i].id+’" >’+status[i].name+'</option>’;
					}
					str += ‘</select>’;
					$(‘select:last’).after(str);
				}
			});
		}
	</script>
</head>
<body>
	<select onchange="change(this)">
	</select>
</body>
</html>
后台:
<?php
	$id = isset($_POST['id']) ? $_POST['id'] : 0 ;
	$conn = mysql_connect('localhost','root','');
	mysql_select_db('mogu');
	$result = mysql_query("set names 'utf8'"); 
	$sql = "select * from city where upid=".$id;
	$result = mysql_query($sql);
	while($rs = mysql_fetch_assoc($result)){
		$arr[] = $rs;
	}
	echo json_encode($arr);
?>
数据库:
http://pan.baidu.com/share/link?shareid=475304&uk=1997302603
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