Simple ajax example using java - problem while receving the response from the servlet

**kolan** · April 13th, 2009 01:37 AM

Hello,

I am trying learn ajax while implementing a simple ajax example using java.

Following is my index.html page where I am trying to use ajax and get some text from the servlet
and display it in the <div>.

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN">
<HTML>
<HEAD>
<TITLE> Ajax Test Page </TITLE>
<script type="text/javascript">

function register()
{
var req;

if (window.XMLHttpRequest)
{
req = new XMLHttpRequest( );
}
else if (window.ActiveXObject)
{
req = new ActiveXObject("Microsoft.XMLHTTP");
}

var url = "register";
req.onreadystatechange = response;
req.open("GET", url, true);
req.send(null);
}

function response()
{
if (req.readyState==4)
{
if (req.status == 200)
{
var time = req.responseText;
alert(time);
document.getElementById("register").innerHTML = req.responseText;
}
}
}
</script>
</HEAD>

<BODY>

<a href="" onClick="register();">Register</a>
<div id="register"> </div>
</BODY>
</HTML>

Following is my servlet where I am sending a text "First Name" in response to the ajax call.

import java.io.*;
import javax.servlet.*;
import javax.servlet.http.*;

public class RegisterForm extends HttpServlet
{
public void doPost(HttpServletRequest request, HttpServletResponse response) throws IOException, ServletException
{
doGet(request, response);
}

public void doGet(HttpServletRequest request, HttpServletResponse response) throws IOException, ServletException
{
response.setContentType("text/html");
response.getWriter().write("First Name");
System.out.println("inside doget");
}
}

When I run this application, I can see that the request is being sent to the servlet (RegisterForm)
"System.out.println("inside doget");" is executed, "inside doget" is printed on to the Tomcat console but the response is not caught in the response() javascirpt method.

Can anyone please tell me where am I going wrong?

**ProgramThis** · April 13th, 2009 08:08 AM

First off, this should probably be in the client side scripting forumn.

However, normally you need to set your response ready state like this:

request.onreadystatechange = function() { response(); };

My guess is that you are never getting into that method and you probably have a javascript error (lower left hand corner of IE) when you try to run this.

At the very top of your response() method set an alert("Hi"); and see if you are even getting there.

**kolan** · April 13th, 2009 02:37 PM

Hello Program This,

First off all, thanks for replying to my post. This is the first time I have ever used a forum. So I am sorry for posting it here.

I followed your suggestions. Modified the code as

req.onreadystatechange = function ()
{
if (req.readyState==4)
{
if (req.status == 200)
{
var time = req.responseText;
alert(time);
document.getElementById("register").innerHTML = req.responseText;
}
}
}

and it worked.

I was following the example at http://www.w3schools.com/Ajax/ajax_source.asp where it says

xmlHttp.onreadystatechange=stateChanged;

for calling the stateChanged function.

Thanks once again.

**ProgramThis** · April 14th, 2009 08:16 AM

That'll teach you a thing about w3schools

Good reference, but not always quite accurate. Glad I could help.

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