สอบถามเรื่อง Ajax หน่อยค่ะ ดูใน Ajax ใน tutor แล้วจับจุดไม่ได้อิอิ

<script>
function fncCheck(){
	var param ="code="+document.getElementById('code').value;		
	var xmlHttp =GetXmlHttpObject();

	 if(xmlHttp== null){
			alert("your browser does not support ");
			return;
	 }
		 
	xmlHttp.onreadystatechange = function(){

	 if(xmlHttp.readyState ==4  && xmlHttp.status==200){
		var  val =xmlHttp.responseText;
		if(val  ==0) alert("ไม่มีค่านี้ในฐานข้อมูล");
		else document.getElementById('showCode').innerHTML=val;
		}
	}
  xmlHttp.open("GET","check.php?"+param,true);
  xmlHttp.send(null);
}
function GetXmlHttpObject(){

var xmlHttp =null;
	if (window.XMLHttpRequest) { // Mozilla, Safari,...
			xmlHttp = new XMLHttpRequest();
			 if (xmlHttp.overrideMimeType) {
				xmlHttp.overrideMimeType('text/html');
			 }
		  } else if (window.ActiveXObject) { // IE
			 try {
				xmlHttp = new ActiveXObject("Msxml2.XMLHTTP");
			 } catch (e) {
				try {
				   xmlHttp = new ActiveXObject("Microsoft.XMLHTTP");
				} catch (e) {}
			 }
		  } 

	return xmlHttp;
}
	</script>
	<input type="text"  id="code" size=15>
	<input type="button" value="Check" onclick="javascript:fncCheck()">
	<div id="showCode"></div>

<?php
//This is code test
// echo "TEST GET HTTP :".$_GET[code];
include ("directory config connect");
$code = $_GET[code]; 
$sql =mysql_query("select * from tbl_name where field_name=$code") or die(mysql_error());
$res= mysql_fetch_array($sql);
$num = mysql_num_rows($sql);
if($num>0){
	echo $res[field_name];
}else{
	echo 0; 
}
?>