filter, forward,ajax call

[Miran Cvenkel]

I have a request to my site, intercept it, and since I know there is nothing, I forward request to valid address

public void doFilter(ServletRequest request, ServletResponse response, FilterChain chain) throws IOException, ServletException 
{
   String url = ((HttpServletRequest)request).getRequestURL().toString();
   if (url.matches("mydomain/gallery/something"))
   {
        //forward to existing address which is mydomain/gallery_old?...
        ((HttpServletRequest) request).getRequestDispatcher("mydomain/gallery_old.jsp?...")).forward(request, response);
         return;
   }     

Then it arrives to the exact page that I want. On that page java script:

   // catchAjaxRequest.jsp is ofcourse in same directory as that page jsp(mydomain/gallery_old.jsp?...) where I just redirected stuff
   doAjaxRequest("catchAjaxRequest.jsp",client_side_catching_function); 

Now the prob is that ajax no longer calls to
catchAjaxRequest.jsp
but instead to
gallery/catchAjaxRequest.jsp
which does not exist there.

Any advice ?

I guess I have to change request, before I forward the whole thing ?

[Rob Spoor]

Moving to Servlets.

[Markas Korotkovas]

In your second snippet you have "catchAjaxRequest.jsp" which is relative path, so when making the request the whole context is included (everything after the domain in your URL). In order to exclude the context add a / like this "/catchAjaxRequest.jsp". This will result in a URL like yourdomain.com/catchAjaxRequest.jsp which is what you are looking for i think.

[Bear Bibeault]

The URL should start with the context path of the web application, which is not necessarily just /.

[Miran Cvenkel]

Thank you both. That was helpful.