AJAX - Request a Server

This example shows how to Sending a request to a server. To send off a request to the server, we use the open() and send() methods.

The open() method takes three arguments. The first argument defines which method to use when sending the request (GET or POST). The second argument specifies the URL of the server-side script. The third argument specifies that the request should be handled asynchronously.

The send() method sends the request off to the server. If we assume that the HTML and PHP file are in the same directory, the code would be:

<html>

<body>



<script type="text/javascript">

function ajaxFunction()

{

var xmlhttp;

if (window.XMLHttpRequest)

  {

  // code for IE7+, Firefox, Chrome, Opera, Safari

  xmlhttp=new XMLHttpRequest();

  }

else if (window.ActiveXObject)

  {

  // code for IE6, IE5

  xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");

  }

else

  {

  alert("Your browser does not support XMLHTTP!");

  }

xmlhttp.onreadystatechange=function()

{

if(xmlhttp.readyState==4)

  {

  document.myForm.time.value=xmlhttp.responseText;

  }

}

xmlhttp.open("GET","time.php",true);

xmlhttp.send(null);

}

</script>



<form name="myForm">

Name: <input type="text" name="username" onkeyup="ajaxFunction();" />

Time: <input type="text" name="time" />

</form>



</body>

</html>
___________________________-
time.php
<?php

response.expires=-1

response.write(time)

?>