2009/7/10 Claus Ibsen <claus.ib...@gmail.com>:
> Hi
>
> Another update on the IN vs OUT when you send in an Exchange.
>
> The ProducerCache that is doing the actual sending when using template
> or sendTo etc, its basically doing all send X to endpoint.
>
> Well I am playing with to let it adhere to the principle Hadrian
> pointed out. He wanted the IN to be more static.
> And we cannot get there yet when you do routing as all the processors
> rely on IN being able to mutate during routing.
>
> Anyway my grief is that when you send in an Exchange the result would
> sometimes be stored on IN and not OUT.
> What I want it to do always is to store the result in OUT and keep IN
> as the original input.
>
> The code to do this is now a bit more complex than just before
>
> // copy the original input
> Message original = exchange.getIn().copy();
>
> producer.process(exchange);
>
> // if no OUT then set current IN as result (except for
> optional out)
> if (!exchange.hasOut() && exchange.getPattern() !=
> ExchangePattern.InOptionalOut) {
> // but only if its not the same as original IN to
> avoid duplicating it
> // and to adhere to the fact that there was no OUT
> result at all
> if (original.getBody() != null &&
> !original.getBody().equals(exchange.getIn().getBody())) {
> exchange.setOut(exchange.getIn());
> }
> }
> // and restore original in
> exchange.setIn(original);
>
> return exchange;
>
>
> What I need to do is to copy the original IN message as it can be
> mutated during routing.
How about we prevent mutation of the IN message? Create a Message
facade which throws UnsupportedOperationException if you try to mutate
it in any way. Then we can pass the same read-only Message around as
the IN within retry loops or from step to step if no new output is
created (e.g. in a content based router where you just move a Message
to the right endpoint without changing it)
--
James
-------
http://macstrac.blogspot.com/
Open Source Integration
http://fusesource.com/
