ตัวอย่างโค้ด ajax_post.php
<script language="javascript" src="ajax.js"></script> <script language="javascript"> function doajax(mydata){ var ajax1=createAjax(); ajax1.onreadystatechange=function(){ if(ajax1.readyState==4 && ajax1.status==200){ document.getElementById('myplace').innerHTML=ajax1.responseText; }else{ return false; } } ajax1.open("POST","data_post.php",true); ajax1.setRequestHeader("Content-Type", "application/x-www-form-urlencoded"); ajax1.send("name="+document.form1.name.value+"&email="+document.form1.email.value); } </script> <form action="" method="post" enctype="multipart/form-data" name="form1" id="form1"> ชื่อ <input name="name" type="text" id="name" /> อีเมลล์ <input name="email" type="text" id="email" /> <input type="button" name="Button" value="Send" onclick="doajax()" /> </form> <p id="myplace"></p>
โค้ด data_post.php
<?php header("Cache-Control: no-store, no-cache, must-revalidate"); header("Cache-Control: post-check=0, pre-check=0", false); if($_POST['name']!=""){ echo $_POST['name']." "; } if($_POST['email']!=""){ echo $_POST['email']; } ?>
ดาวน์โหลด ajax.js คลิกดาวน์โหลด
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