ServletContext aContext = servlet.getServletContext(); System.out.println("aurl in encodeforward is: " + aURL); RequestDispatcher aDispatch = aContext.getRequestDispatcher(res.encodeURL(aURL));
if (aDispatch == null) { System.out.println("Request dispatcher could not be gotten..."); } aDispatch.forward(req, res); } Please can anyone throw me any light on this or let me know of any other way to do this. Thanks, Deepthi
maha anna
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Joined: Jan 31, 2000
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Deepthi, Please refer to the sample code in this thread. Here the forwarded resource could be anything html/jsp/servlet. http://www.javaranch.com/ubb/Forum7/HTML/002833.html regds maha anna [This message has been edited by maha anna (edited March 30, 2001).]