Spring Integration Transform 失败回滚 JMS 并且不转发到错误通道

Question

使用 Boot 2.2.2 和 Integration 5.2.2 - 当 XML 消息来自 File 并且解组失败（即它不是 XML）时，消息会按预期继续到 errorChannel。但是，当消息来自 JMS，通过相同的通道路由并且解组失败时，它不路由到 errorChannel，并且消息回滚到 JMS。之后，对于同一条消息，我陷入了SAXParseException 的无限循环。

我从 Proper ultimate way to migrate JMS event listening to Spring Integration with Spring Boot 开始关注这个例子。是否有一些我没有考虑的隐含交易控制？如何让 Spring Integration 将消息转发到 errorChannel 并从传入队列提交“get”？

代码概要如下；

    @Bean
    public IntegrationFlow fileReader() {
        return IntegrationFlows
            .from(
                Files
                    .inboundAdapter( ... )
                    ...
                    .get(), e -> e.poller(Pollers.fixedDelay(1000))
            )
            .transform(new FileToStringTransformer())
            .channel("backUpChannel")
            .get();
    }

    @Bean
    public IntegrationFlow getMessageFromJms(ConnectionFactory connectionFactory, @Value("${queues.myQueue}") String myQueue) {
        return IntegrationFlows.from(
            Jms
                .messageDrivenChannelAdapter(connectionFactory)
                .destination(myQueue)
        )
        .channel("backUpChannel")
        .get();
    }

    @Bean
    public IntegrationFlow doBackUp() {
        return IntegrationFlows
            .from("backUpChannel")
            .<String>handle((payload, headers) -> {
                String uuid = headers.get(MessageHeaders.ID).toString();
                File backUpFile = new File("c:/backup/" + uuid + ".txt");
                byte[] payloadContent = payload.getBytes();
                try {
                    java.nio.file.Files.write(backUpFile.toPath(), payloadContent);
                } catch (IOException e) {
                    e.printStackTrace();
                }
                return payload;
            })
            .channel("XXX")
            .get();
    }

    @Bean
    public Jaxb2Marshaller unmarshaller() {
        Jaxb2Marshaller unmarshaller = new Jaxb2Marshaller();
        unmarshaller.setClassesToBeBound(MyClass.class);
        return unmarshaller;
    }

    @Bean
    public IntegrationFlow handleParseXml() {

        return IntegrationFlows
            .from("XXX")
            .transform(new UnmarshallingTransformer(unmarshaller()))
            .channel("YYY")
            .get();
    }

Answer 1

您需要将.errorChannel(...) 添加到消息驱动的通道适配器中。