اینم کد:
page:"ajax.js"
var http=getHttp();
var code=0;
//***********************
function getHttp()
  {
var xmlhttp;
  try
    {
       xmlhttp=new ActiveXObject("Msxml2.XMLHTTP");  
     }
  catch (e)
    {
    // Internet Explorer
    try
      {
      xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
      }
      catch (e)
      {
        if(typeof XMLHttpRequest !='undefiend')
         {
           xmlhttp=new XMLHttpRequest();
         }
      }
    
     }
     
 
   return xmlhttp;
   } 
//*******************************
function sendtorecieveforadd(){
title=document.getElementById("tftitle").value; 
agency=document.getElementById("tfagency").value;
 items=title+","+agency;  
 
 http.open("GET","new.php?params="+items,false);
 http.onreadystatechange=answerofadd;
 http.send(null);
}
//************************
function answerofadd()
{
 if(http.readystate==4)
   {
     r=http.responseText;
     
     newsid=r;
  if(r!="-1")
    
     Add(newsid,title,agency);
 }      
     
}
//******page index********
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Language" content="en-us" />
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Title</title>
<style type="text/css">
.style1 {
    text-align: center;
}
</style>
<script type="text/javascript" src="ajax.js" defer="defer"></script>
</head>
<body >
Title:<input name="tftitle" type="text" />
<p>Agency:<input name="tfagency" type="text" />
<input name="Button1" type="submit" value="Add"  onclick="sendtorecieveforadd();"/>
</body>
</html>