/*Below Spring-integration configuration*/
/*Listener*/
<!--Reading the Message from Queue as a Listener -->
<int-jms:inbound-gateway connection-factory="MQConnectionFactory"
request-destination="ReadWsRequestQueue"
request-channel="ReadWsInputChannel"
transaction-manager="hibernateTransactionManager"
error-channel="errorReadChannel"/>
/*Processing Message*/
<!--Processing the message-->
<int:chain input-channel="ReadWsInputChannel">
<int:transformer ref="ReadUnmarshaller"/>
<int:transformer ref="RequestBuilder" method="build"/>
<int:service-activator ref="ReadService" method="registerRead" />
<int:transformer ref="ResponseBuilder" method="buildResponse"/>
<int:transformer ref="ReadMarshaller"/>
<int:transformer ref="toStringTransformer"/>
/*Error Channel config*/
<!--Error channel configuration -->
<int:channel id="errorReadChannel"/>
<int:chain input-channel="errorReadChannel">
<int-jms:outbound-gateway id="jmsOutboundGateway"
connection-factory="MQConnectionFactory"
request-destination="DLQErrorQueue" />
</int:chain>
问题:
在服务Activator方法中,我们抛出一个RuntimeException,它应该被重定向到错误通道?
所有例外都应该转到错误频道
问题2。)
在Spring集成中是否有任何方法可以将异常转发到单独的通道?