Somehow below code worked,  I thought it be helpful for someone else to.

if(httpRequest.getSession(false) == null){
                Enumeration headerNames = httpRequest.getHeaderNames();        
                while(headerNames.hasMoreElements()) {
                  String headerName = (String)headerNames.nextElement();
                  if(headerName.equalsIgnoreCase("Wicket-Ajax")){
                          PrintWriter out = httpResponse.getWriter();        
                          out.println("<?xml version='1.0' 
encoding='UTF-8'?><ajax-response><header-contribution encoding='wicket1' 
><![CDATA[<head xmlns:wicket='http://wicket.apache.org'>");
                          
out.println("</head>]]></header-contribution><evaluate><![CDATA[ 
location.href='" +  httpRequest.getContextPath() + 
"';]]></evaluate></ajax-response>");
                         return;
                         
                  }
               
                }
       } 

-----Original Message-----
From: gurpreet.si...@wellsfargo.com [mailto:gurpreet.si...@wellsfargo.com] 
Sent: Thursday, July 22, 2010 12:49 PM
To: users@wicket.apache.org
Subject: Session time out for Ajax requests

Hi All,

Any one tried this :

"In your served filter you will have to check for Wicket-Ajax header
and if it is present return a properly formatted Ajax-response that
contains JavaScript to redirect to your login page.


I am not sure what exactly I need to return. I tried with

<ajax-reponse><script 
language="javascript">location.href="mypage"</script></ajax-response>

And
<ajax-response><![CDATA[<script 
langauge='javascript'>location.href="mypage"</script>]]</ajax-response>



But nothing seem to be working. Any idea what's wrong here

Thanks in Advance.





---------------------------------------------------------------------
To unsubscribe, e-mail: users-unsubscr...@wicket.apache.org
For additional commands, e-mail: users-h...@wicket.apache.org

Reply via email to