I don't think it works. My exact route is:
from("jetty:http://localhost:9080/myapp/myservice";)
.process(new Processor() {...})
.to("jms:queue:some_queue")
.to("http://localhost:9080/myapp/myservice";);
I still got the error said,
org.apache.camel.ExchangeTimedOutException: The OUT message was not received
within: 20000 millis on the exchange:
Basically I would like my starting endpoint receive 200 status ok when send
jms successfully or 500 error when send jms failed. How to do that?
Thanks a lot.
Claus Ibsen-2 wrote:
>
> On Wed, Aug 12, 2009 at 7:01 PM, janylj<[email protected]> wrote:
>>
>> Thanks a lot, Willem. It works using inOnly()
>>
>> willem.jiang wrote:
>>>
>>> from("jetty ....").process(...).inOnly().to("jms:queue")
>>>
>>
>> How to make it send a HttpResponse only if sending JMS message is
>> successfully. I tried to list the same jetty endpoint in the end, but it
>> doesn't work. Please advise. Thanks again.
>>
>> from("jetty ....").process(...).to("jms:queue").to("jetty....")
>>
>
> jetty is only used as a consumer, eg you can only use it in the from node.
> So when you want to call some external HTTP service use the http
> component instead
>
> from(jetty).to(http) = OK
>
> from(jetty).to(jetty) = NO OK
>
>
>> --
>> View this message in context:
>> http://www.nabble.com/What-to-do-with-HTTP-response-when-route-from-HTTP-to-JMS--tp24924548p24940755.html
>> Sent from the Camel - Users mailing list archive at Nabble.com.
>>
>>
>
>
>
> --
> Claus Ibsen
> Apache Camel Committer
>
> Open Source Integration: http://fusesource.com
> Blog: http://davsclaus.blogspot.com/
> Twitter: http://twitter.com/davsclaus
>
>
--
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