"Kommana, Sridhar" wrote:
>
> We are using IBMs WebSphere. I am using RequestDispatcher to redirect the servlet to
>a different page/servlet. Code I am using in my
> servlet(x.java) is as follows:
>
> String xyz = request.getParameter("xyz");
>
> if (xyz.equals("one")) {
> RequestDispatcher rd1 =
>getServletContext().getRequestDispatcher(page1.jsp);
> rd1.forward(request, response);
> }
>
> RequestDispatcher rd2 =
>getServletContext().getRequestDispatcher(page2.htm);
> rd2.forward(request, response);
>
> If "xyz" is not "one" page2.htm is displayed. If "xyz" is "one", then page1.jsp is
>displayed and "llegalStateException" error is
> thrown below it. I am not sure why it is behaving like this. I believe if "xyz" is
>"one" then control should be transferred to
> page1.jsp and code after the if loop should not be executed.
>
> I have tried using servlets in place of page1 and page2. The result is same - code
>upto forward takes the control to the target
> servlet and any code after the forward including forward if any, are also executed.
>If there is no code after forward(esp another
> forward) then it behaves just fine.
>
> Can anybody throw light on it!
You have to add a "return;" after the forward() call. There's no magic in Java
that
let's it figure out by itself that it should not continue when the forward()
call
returns.
In a JSP page, using <jsp:forward>, there *is* some magic though; the
<jsp:forward>
tag tells the JSP container that the rest of the page should be ignored (using a
special return code, see the JSP 1.1 spec for details). But if you forward with
regular Java code, you have to be your own magician.
Hans
--
Hans Bergsten [EMAIL PROTECTED]
Gefion Software http://www.gefionsoftware.com
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