Rob_Mathews_at_Dell.com wrote:
>
> Hi:
>
> I’ve read the posts and noticed a lack of Jgroups support. The closest
> it seems to come is a mention of reimplementing the SPI layer. I
> grabbed the latest source from CVS and there is no JGroups
> implementation yet. The JXTA implementation of that looks complex –
> runs to some 7000 lines, and probably represents a 6 month effort to
> reimplement in JGroups. This looks bleak.
>
> Is there some other way that I’m missing?
>
> I should mention that I’m be pushed this way by these posts, which
> talk about be unable to do test driven development:
>
> The testing problem:
> http://forums.java.net/jive/thread.jspa?messageID=330040&tstart=0
> <http://forums.java.net/jive/thread.jspa?messageID=330040&tstart=0>
>
JXTA ships with a utility (NetworkManager) allowing developers to easily
start and stop JXTA. This utility has evolved under Shoal to support
multi infrastructure group participation, at some it will be back
ported, as demand for it in the JXTA platform grows.
> a very good blog entry: JXTA: Not The Solution To Java Peer Discovery
> <http://www.prestonlee.com/2007/04/14/jxta-not-the-solution-to-java-peer-discovery/101/>
>
This Blog is mainly about the misuse of the PeerDiscovery protocol to
discover all participants in a given group, the discovery protocol is
meant to bootstrap, not an exhaustive search, that's like saying
enumerate all computers on the internet. Shoal uses other JXTA protocols
to achieve it's discovery of cluster members.
> At a minimum, my project has two requirements that are driving me
> towards JGroups:
>
> - A design that requires for multiple members of a group within the
> same JVM (same Tomcat instance, actually)
>
You should be able to do so today.
> - A testing requirement that requires multiple members of a group
> within different processes on the same machine (ie, different Tomcat
> instances on the same machine listening on different ports)
>
Ditto.
> I’m under the impression that JXTA prevents me from doing either of
> those things.
>
> Is that right?
>
> Thanks,
>
> Rob.
>