Re: Peer-Discovery in own Group
Bernard Traversat <[email protected]>
| Newsgroups | gmane.comp.java.jxta.user |
|---|---|
| Message-ID | <[email protected]> |
Volker Katz wrote: > On Wednesday 27 July 2005 13:54, Joachim Goetze wrote: > >>Volker Katz wrote: >> >>>Now I want to discover my neighbours in this group. At the moment, >>>I do this job like in programmer's guide with my custom >>>DiscoveryListener. There I search for neighbours by calling: >>>discovery.getRemoteAdvertisements(null, DiscoveryService.PEER,null, >>>null, 5, this); >>>discovery points to discovery = myGrp.getDiscoveryService(); >>> >>>But I never get an DiscoveryEvent! If I change discovery to >>>netpeergroup I get discoveryEvents... > > >>You should ensure that you are connected to a rendezvouz peer. >>Without that connection you will never receive any discovery events. > > > thank you very much for your hints. I'll test our code right now. But > just for understandig: > I know, I have to be connected to rendezvous-peer. I'm searching for > peers in my peergroup (a subgroup of netgroup). So, do I need to > another rendezvous-peer for my g Yes, every peergroup defines its own network context and uses its own set of rendezvous. You can use two main policies either force one of your peer to act as RDV for the subgroup, or use the auto-elect RDV feature to let peers self-elect as RDV when there is a need to. B. > Because it's my group and I have > to create a rendezvous-peer myself, haven't I? > > Thanks & best regards! > Volker > > --------------------------------------------------------------------- > To unsubscribe, e-mail: [email protected] > For additional commands, e-mail: [email protected] > -- --http://weblogs.java.net/blog/tra "As Java implies platform independence, and XML implies language independence, JXTA implies network independence."