On Wed, 19 Jan 2005 22:16:34 -0800 Dakota Jack <[EMAIL PROTECTED]> wrote:
<snip>
> 2) Specify a <forward /> for the Action that sends the user back to
> from whence they came.
I don't think I can do this because they may come from a number of pages.
</snip>
You can specify a dynamic ActionForward return in your Action class execute(...) method. Not sure why this presents any difficulty at all.
Jack
My apologies for not reading first - yeah - use a dynamic one like jack said. I would PROBABLY subclass Action with an abstract subclass - something like:
public ... execute(...) {
// This overrides Action.execute()// Do whatever up-front checking
...
// Now, call my own REAL execute method:
forward = this.doWork(mapping, form, request, response);
// Store the path of the forward in the user's session
// so that on their next communication back to us,
// the go right back to the same place
request.getSession().setAttribute("i_was_here", forward.getPath());
}
// Here's a method for your doWork() implementation to
// just return a new forward for where they were before.
// You don't HAVE to call this, of course.
protected ActionMapping getDefaultForward(mapping, request) {
try {
return new ActionForward((String)request.getSession(true).getAttribute("i_was_here"));
} catch (Excetpion e) {
// Could be an NPE (attribute doesn't exist)
// or a ClassCastException (attribute doesn't cast to a String)
return mapping.findForward("default");
}
}
public abstract void doWork(ActionMapping mapping, ActionForm form, HttpServletRequest request, HttpServletResponse response) throws Exception;
Do NOT just copy and paste that - I've not tried it. But rather than use the Referer header, I think that's a more sensible solution. When you call an action of this class, it remembers where it forwards you to. Then, when you call another subclass of this, if you want it to send you right back where you were, your doWork() implementation will just return getDefaultForward(mapping, request).
w
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