java - 将消息从 ServiceActivator 传递到新的 InboundChannelAdapter

Question

我已经实现了Spring(启动)集成，我有一个InboundChannelAdapter，这个方法通过RestTemplate发送Web请求，并将结果传递给Splitter、Filter，最后到一个ServiceActivator，ServiceActivator将数据保存到数据库中。我想将消息传递给新的“InboundChannelAdapter”，以便在将数据保存到数据库后对消息进行一些额外的处理，我应该做什么才能实现这个？
我的解决方案是正确的解决方案吗？

@InboundChannelAdapter(value = "channel1", poller = @Poller("downloadTrigger"))
public ResponseEntity<AppsItemParser[]> download()
{
    String url = config.getUrl();
    // Call a rest webservice
    return responseEntity;
}

@Splitter(inputChannel = "channel1", outputChannel = "channel2")
public List<AppsItemParser> scrape(ResponseEntity<AppsItemParser[]> payload)
{
    return new ArrayList<AppsItemParser>(Arrays.asList(payload.getBody()));
}

@Transformer(inputChannel = "channel2", outputChannel = "channel3")
public App convert(AppsItemParser payload)
{
    App app = new App();
    app.setAddDate(payload.getAddDate());
    app.setSize(payload.getSize());

    return app;
}

@Filter(inputChannel = "channel3", outputChannel = "channel4")
public boolean filter(App app)
{
    final Set<ConstraintViolation<App>> violations = validator.validate(app);
    if (violations != null && !violations.isEmpty())
    {
        return false;
    }
    return true;
}

  @ServiceActivator(inputChannel = "channel4")
    public void save(App app)
    {
        appRepository.save(app);
        //Send message to another spring integration flow
    }

Answer 1

入站 channel 适配器定义流的开始；它们不接收来自其他组件的消息；相反，让您的服务返回结果...

@ServiceActivator(inputChannel = "channel4", outputChannel="next")
public App save(App app)
{
    appRepository.save(app);
    return app;
}

或

@ServiceActivator(inputChannel = "channel4", outputChannel="next")
public SomeResult save(App app)
{
    appRepository.save(app);
    return new SomeResult(...);
}