java - 如何强制Rabbit MQ重新累积并发送消息？

Question

我在 XML 文件中配置了几个 Spring-Integration 元素(见下文) 从 amqp channel 适配器，消息被定向到具有实现integrationRouterImpl 的路由器integrationSecondaryRouter。

如果在integrationRouterImpl中存在未捕获的异常，我希望Rabbit MQ会一次又一次地发送消息。然而，这并没有发生。 Rabbit MQ 监视器不显示任何消息累积。我的配置有错误吗？

<int-amqp:inbound-channel-adapter 
    channel="integrationFrontDoorQueueChannel" 
    queue-names="${integration.creation.orders.queue.name}" 
    header-mapper="integrationHeaderMapper"
    connection-factory="connectionFactory" 
    error-channel="errorChannel" 
/>

<int:chain 
        id="integrationFrontDoorQueueChain" 
        input-channel="integrationFrontDoorQueueChannel"
        output-channel="integrationRouterChannel">
    <int:transformer ref="integrationJsonPayloadTransformer" method="transformMessagePayload"/>
    <int:filter ref="integrationNonDigitalCancellationFilter" method="filter"/>
    <int:filter ref="integrationPartnerFilter" method="filter"/>
    <int:filter ref="integrationOrderDtoDgcAndGoSelectFilter" method="filter"/>

</int:chain>

 <int:header-value-router 
    id="integrationPrimaryRouter"
    input-channel="integrationRouterChannel" 
    default-output-channel="integrationFrontDoorRouterChannel"
    resolution-required="false"
    header-name="#{T(com.smartdestinations.constants.SdiConstants).INTEGRATION_PAYLOAD_ACTION_HEADER_KEY}">
    <int:mapping 
        value="#{T(com.smartdestinations.service.integration.dto.IntegrationAction).EXCLUSION_SCAN.name()}" 
        channel="integrationExclusionChannel" 
    />
</int:header-value-router>


<int:router 
        id="integrationSecondaryRouter"
        ref="integrationRouterImpl" 
        input-channel="integrationFrontDoorRouterChannel"
        method="route" 
        resolution-required="false" 
        default-output-channel="nullChannel"
/>

Answer 1

看，您有 error-channel="errorChannel" 和 Documentation就此事指出:

The default "errorChannel" is a PublishSubscribeChannel.

是的，有一名订阅者。但它只是_org.springframework.integration.errorLogger。

由于没有人将您的异常重新抛出到 SimpleMessageListenerContainer，因此没有理由 nack 消息并再次重新传递它。