从春季整合中收集者回来的反应不一致

Question

我正在使用散点机图案，并拨打3个子流程。然后收集它们并汇总它们。我试图在汇总后取回结果。我将响应作为列表，但是列表中的两个响应并不一致。当我试图通过执行（0）来获得。这是不一致的。请建议我一种理解输出的方法，以便我可以将其分配给实际变量。

配置文件

 @Configuration
    public class IntegrationConfiguration {
      @Autowired LoansServiceImpl loansService;
    
      long dbId = new SequenceGenerator().nextId();
  //   Main flow
  @Bean
  public IntegrationFlow flow() {
    return flow ->
        flow.split()
            .log()
            .channel(c -> c.executor(Executors.newCachedThreadPool()))
            .convert(LoanProvisionRequest.class)
            .scatterGather(
                scatterer ->
                    scatterer
                        .applySequence(true)
                        .recipientFlow(flow1())
                        .recipientFlow(flow2())
                        .recipientFlow(flow3()),
                gatherer -> gatherer.releaseLockBeforeSend(true))
            .log()
            .aggregate(a -> a.outputProcessor(MessageGroup::getMessages))
            .channel("output-flow");
  }
  //   flow1
  @Bean
  public IntegrationFlow flow1() {
    return integrationFlowDefination ->
        integrationFlowDefination
            .channel(c -> c.executor(Executors.newCachedThreadPool()))
            .handle(
                message -> {
                  try {
                    lionService.saveLionRequest(
                        (LionRequest) message.getPayload(), String.valueOf(dbId));
                  } catch (JsonProcessingException e) {
                    throw new RuntimeException(e);
                  }
                });
  }

  //   flow2
  @Bean
  public IntegrationFlow flow2() {
    return integrationFlowDefination ->
        integrationFlowDefination
            .channel(c -> c.executor(Executors.newCachedThreadPool()))
            .handle(
                message ->
                    lionService.getData(
                        (LionRequest) message.getPayload(), SourceSystem.PROVISION))
            .log();
  }

  //  flow3
  @Bean
  public IntegrationFlow flow3() {
    return integrationFlowDefination ->
        integrationFlowDefination
            .channel(c -> c.executor(Executors.newCachedThreadPool()))
            .handle(
                message ->
                    lionService.prepareCDRequest(
                        (LionRequest) message));
  }

  @Bean
  public MessageChannel replyChannel() {
    return MessageChannels.executor("output-flow", outputExecutor()).get();
  }

  @Bean
  public ThreadPoolTaskExecutor outputExecutor() {
    ThreadPoolTaskExecutor pool = new ThreadPoolTaskExecutor();
    pool.setCorePoolSize(4);
    pool.setMaxPoolSize(4);
    return pool;
  }
}

网关

@MessagingGateway
public interface LionGateway {

  @Gateway(requestChannel = "flow.input", replyChannel = "output-flow")
  List<?> echo(LionRequest lionRequest);
}

控制器

@Autowired private LionGateway lionGateway;

 @PostMapping(value = "/invoke-integration")
  public String invokeIntegrationFlow(@RequestBody LionRequest lionRequest) {
    String response1 = lionGateway.echo(lionRequest).get(0);
    String response2 = lionGateway.echo(lionRequest).get(1);
    System.out.Println("response2 ")
    System.out.Println("response1 ")

    return "response";
  }

控制器中的 I有两个变量，即响应1＆amp;响应2这两个值是不一致的。用更好的方法帮助我。

Answer 1

由于您使用的是执行人频道，因此订单仅仅是答复返回的顺序。收集者对下游拓扑一无所知。

您需要处理代码中回复列表的随机顺序。