อยากทำ ajax แบบ drop down ที่เวลาเลือกใน drop down แล้ว ให้แสดงข้อมูลบนหน้านั้นเลยคะ

<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<select id="selMember">
	<?php
    	$select=mysql_query("select * from member") or die (mysql_error());
		while($result=mysql_fetch_array($select)){
	?>
    <option value="<?=$result['MemberID'];?>"><?=$result['Member'];?></option>
    <? } ?>
</select>
<div id="detail"></div>

<script src="admin/bootstrap/js/jquery-1.10.2.min.js"></script> // download www.jquery.com
<script>
	$(function(){
		$('#selMember').change(function(){
			var detail=$('#detail');
			detail.empty();
			$.ajax({url:"check.php",
				async:false,
				cache:false,
				global:false,
				type:"POST",
				data:"MemberID="+$(this).val(),
				success: function(result){
					var obj=jQuery.parseJSON(result);
					$.each(obj,function(key,val){
						detail.append(val['MemberID']);
						detail.append(val['Member']);
						detail.append(val['Address']); // fields จากฐานข้อมูล
					});	
				}
			});	
		});
	});
</script>

<?
	include ("config.inc.php");
	$MemberID=$_POST['MemberID'];
	$arrResult=array();
	$select=mysql_query("select * from member WHERE MemberID='".$MemberID."'") or die (mysql_error());
	$numFields=mysql_num_fields($select);
	while($result=mysql_fetch_array($select)){
		$arrCol=array();
		for($i=0;$i<$numFields;$i++){
			$arrCol[mysql_field_name($select,$i)]=$result[$i];
		}
		array_push($arrResult,$arrCol);
	}
	echo json_encode($arrResult);
?>