You can return a StreamResponse whose job is to generate an
InputStream of XML to be sent to the client.
On Thu, Apr 23, 2009 at 6:12 AM, Borut Bolčina <borut.bolc...@gmail.com> wrote:
> Hello,
>
> I have a dispatcher which matches some URL pattern like this:
>
> public boolean dispatch(Request request, Response response) throws
> IOException {
> String path = request.getPath();
> // just imagine this works ok
> String uglyWord = request.getContextPath() == "" ?
> StringUtils.substringAfterLast(path, "/") : path.substring(1);
>
> if (uglyWordSet.contains(uglyWord)) {
> // return some XML if url contains some ugly word
> // WHAT IS THE BEST APPROACH TO RETURN XML (STREAM) ?
> return true;
> }
>
> return false;
> }
>
> I know how to create a page which returns xml stream response (I
> actually have one). Can I just reuse this page?
>
> Using T 5.0.18.
>
> Regards,
> Borut
>
> ---------------------------------------------------------------------
> To unsubscribe, e-mail: users-unsubscr...@tapestry.apache.org
> For additional commands, e-mail: users-h...@tapestry.apache.org
>
>
--
Howard M. Lewis Ship
Creator of Apache Tapestry
Director of Open Source Technology at Formos
---------------------------------------------------------------------
To unsubscribe, e-mail: users-unsubscr...@tapestry.apache.org
For additional commands, e-mail: users-h...@tapestry.apache.org
