Re: [JXTA discuss] How do I manually start a resolver service

If your service relies on other services to be available, then it should 
return START_AGAIN_STALLED to indicate to the group that it should be 
called again, it should return START_AGAIN_STALLED until the required 
services have started, after which it should return START_OK.  The 
service should also postpone any calls to the required service until 
such services have started.

Also the ResolverService has always been part the JXTA group spec 
definition (1.0 through 2.x), and have always been started unless 
explicitly removed from a group definition.

Mohamed



    public int startApp(String[] arg) {
        resolver = group.getResolverService();

        if (null == resolver) {
            if (LOG.isEnabledFor(Level.WARN)) {
                LOG.warn("Stalled until the resolver service");
            }
            return Module.START_AGAIN_STALLED;
        }



[email protected] wrote:

> Hi,
>  
> I have created a JXTA service and added it to a new peergroup and 
> started that peergroup. Then, I successfully lookupService(new 
> service). The purpose of the new services is to pass messages between 
> two peers. 
>  
> However, message are lost. I guess that the resolver services is not 
> started on the new peergroup that I created. see message.
>  
> <WARN  2005-03-30 15:49:21,152 
> DiscoveryServiceImpl::getRemoteAdvertisements:291> resolver has not 
> started yet, query discarded.
>  
> <WARN  2005-03-30 15:58:40,788 ResolverServiceImpl::processSrdi:1051> 
> No srdi handler registered 
> :urn:jxta:uuid-DEADBEEFDEAFBABAFEEDBABE0000000305
>  
> But, I don't know how to manually started the ResolverService on a new 
> peergroup. Or, that may not be the problem. Does anyone have good 
> ideas or explainations for it.
>  
> Thanks,
>  
> -Ray
>  
>