我遇到错误处理问题。如果jrnRoute
中发生异常,则由exceptionHandler
处理器处理。还行吧。但导致异常的消息仍然保留在jrnQueue
中,再次处理并导致错误一遍又一遍。我也在日志中发出警告(见下文)。如何阻止无限重新传递?我想在发生错误时抛出一条消息。
Camel configuration
<camel:onException>
<camel:exception>java.lang.Exception</camel:exception>
<camel:redeliveryPolicy disableRedelivery="true" />
<camel:process ref="exceptionHandler" />
<camel:rollback markRollbackOnly="true" />
</camel:onException>
<camel:route id="translatorRoute">
<camel:from ref="transactionsQueue" />
<camel:process ref="messageTranslator" />
<camel:inOnly ref="apfRequestQueue" />
</camel:route>
<camel:route id="jrnRoute">
<camel:from ref="jrnQueue" />
<camel:process ref="jrnProcessor" />
<camel:stop />
</camel:route>
</camel:camelContext>
Warning
11:15:00,141 WARN [JmsMessageListenerContainer] Execution of JMS message listener failed, and no ErrorHandler has been set.
org.apache.camel.RuntimeCamelException: org.apache.camel.RollbackExchangeException: Intended rollback. Exchange[JmsMessage: [ObjectMessageImpl com.swiftmq.jms.ObjectMessageImpl@c84d9d
messageIndex = 5_2
messageId = [LazyUTF8String, s=ID:/172.26.214.11/5349789614428334512/10/0, buffer=[B@5f7fd8]
userId = [LazyUTF8String, s=null, buffer=[B@1c254aa]
clientId = null
timeStamp = 1316682898526
correlationId = null
replyTo = null
destination = jrnQueue@z4smq_4001
deliveryMode = 2
redelivered = true
deliveryCount = 8
type = null
expiration = 0
priority = 4
props = {...}
readOnly = true
sourceRouter = null
destRouter = null
destQueue = null array=[B@1447e6b cnt=1295]]
at org.apache.camel.util.ObjectHelper.wrapRuntimeCamelException(ObjectHelper.java:1145)
at org.apache.camel.component.jms.EndpointMessageListener.onMessage(EndpointMessageListener.java:108)
at org.springframework.jms.listener.AbstractMessageListenerContainer.doInvokeListener(AbstractMessageListenerContainer.java:560)
at org.springframework.jms.listener.AbstractMessageListenerContainer.invokeListener(AbstractMessageListenerContainer.java:498)
at org.springframework.jms.listener.AbstractMessageListenerContainer.doExecuteListener(AbstractMessageListenerContainer.java:467)
at org.springframework.jms.listener.AbstractPollingMessageListenerContainer.doReceiveAndExecute(AbstractPollingMessageListenerContainer.java:325)
at org.springframework.jms.listener.AbstractPollingMessageListenerContainer.receiveAndExecute(AbstractPollingMessageListenerContainer.java:243)
at org.springframework.jms.listener.DefaultMessageListenerContainer$AsyncMessageListenerInvoker.invokeListener(DefaultMessageListenerContainer.java:1058)
at org.springframework.jms.listener.DefaultMessageListenerContainer$AsyncMessageListenerInvoker.executeOngoingLoop(DefaultMessageListenerContainer.java:1050)
at org.springframework.jms.listener.DefaultMessageListenerContainer$AsyncMessageListenerInvoker.run(DefaultMessageListenerContainer.java:947)
at java.util.concurrent.ThreadPoolExecutor$Worker.runTask(ThreadPoolExecutor.java:886)
at java.util.concurrent.ThreadPoolExecutor$Worker.run(ThreadPoolExecutor.java:908)
at java.lang.Thread.run(Thread.java:662)
Caused by: org.apache.camel.RollbackExchangeException: Intended rollback. Exchange[JmsMessage: [ObjectMessageImpl com.swiftmq.jms.ObjectMessageImpl@c84d9d
messageIndex = 5_2
messageId = [LazyUTF8String, s=ID:/172.26.214.11/5349789614428334512/10/0, buffer=[B@5f7fd8]
userId = [LazyUTF8String, s=null, buffer=[B@1c254aa]
clientId = null
timeStamp = 1316682898526
correlationId = null
replyTo = null
destination = jrnQueue@z4smq_4001
deliveryMode = 2
redelivered = true
deliveryCount = 8
type = null
expiration = 0
priority = 4
props = {...}
readOnly = true
sourceRouter = null
destRouter = null
destQueue = null array=[B@1447e6b cnt=1295]]
答案 0 :(得分:3)
是的markRollbackOnly会导致事务管理器将TX标记为回滚,因此JMS代理会将消息保留在队列中,并再次重新发送。所以删除它。
并按照上面的Ben发布,处理为true:
<camel:onException>
<camel:exception>java.lang.Exception</camel:exception>
<camel:redeliveryPolicy disableRedelivery="true" />
<camel:handled>
<camel:constant>true</camel:constant>
</camel:handled>
<camel:process ref="exceptionHandler" />
</camel:onException>
答案 1 :(得分:2)
看起来markRollbackOnly="true"
就是问题......
我通常会做这样的事情来简单处理(防止传播回调用者),防止重试并记录错误消息......
onException(Exception.class)
.handled(true).maximumRedeliveries(0)
.to("log:error processing message");