在Spring Integration中，DSL如何指定对已发布频道的订阅？

Question

当使用Spring Integration DSL builder模式时，它通常会“自动”填充元素之间所需的通道。然而，有时情况并非如此。

在高层次上，包装应用程序将元数据保存在数据库中，以便根据需要动态创建和销毁我们（可能）从未见过的跨平台流。因此，流不适合使用静态符号（如@Bean）实例化，而是必须动态创建和销毁，并在运行时在Spring上下文中注册/取消注册。

我有一个在动态创建的主流中使用的已发布消息通道和一个在动态创建的子流中使用的通道，但是我看不到如何从子流订阅main PublishChannel。

这让我把消息推入频道，但没有订阅，什么都不会发生。

提前感谢。

一些先前的研究（不是详尽的清单：

https://github.com/spring-projects/spring-integration-flow

https://dzone.com/articles/spring-integration-building

https://xpadro.com/2014/05/spring-integration-4-0-a-complete-xml-free-example.html

Spring集成网关"Dispatcher没有订阅者"

Spring集成-如何调试“Dispatcher没有订阅服务器”？

原木剪

task-scheduler-1 2020-12-31 00:25:32,526 INFO  o.s.i.g.GatewayProxyFactoryBean - started b653ca1c-038d-4567-bd4e-4c16ecc502a3.org.springframework.integration.config.ConsumerEndpointFactoryBean#3#gpfb
task-scheduler-1 2020-12-31 00:25:32,538 DEBUG o.s.i.c.PublishSubscribeChannel - preSend on channel 'b653ca1c-038d-4567-bd4e-4c16ecc502a3.mainPublishChannel', message: GenericMessage [payload=[{... timestamp=1609395932538}]
task-scheduler-1 2020-12-31 00:25:32,539 DEBUG o.s.i.d.BroadcastingDispatcher - No subscribers, default behavior is ignore
task-scheduler-1 2020-12-31 00:25:32,539 DEBUG o.s.i.c.PublishSubscribeChannel - postSend (sent=true) on channel 'b653ca1c-038d-4567-bd4e-4c16ecc502a3.mainPublishChannel', message: GenericMessage [payload=[{aaa=ee}], headers={aaa=ee, sequenceNumber=1, replyChannel=org.springframework.messaging.core.GenericMessagingTemplate$TemporaryReplyChannel@7771e96a, errorChannel=org.springframework.messaging.core.GenericMessagingTemplate$TemporaryReplyChannel@7771e96a, sequenceSize=2, yyy=2020-12-24 11:15:30.915278, correlationId=0eef0e4e-768c-90db-fa7b-2d1767335b26, timestamp=1609395932538}]

代码剪切：

    String channelId=getId().toString()+'.'+"mainPublishChannel";
    MessageChannel channel = MessageChannels.publishSubscribe(channelId, stepTaskExecutor).get();

    final IntegrationFlowBuilder bldr = IntegrationFlows
        .from(setupAdapter,
                c -> c.poller(Pollers.fixedRate(pollerFixedRate, TimeUnit.MILLISECONDS).maxMessagesPerPoll(1)))
        .enrichHeaders(h -> h.headerExpression("xxx", "payload[0].get(\"xxx\")")
                .headerExpression("yyy", "payload[0].get(\"yyy\")"))
        .split(tableSplitter)
        .gateway(channel)   
        .routeToRecipients(r -> this.buildRecipientListRouterSpecForRules(r, rules, channel)
                    ) 
        .aggregate()
        .handle(cleanupAdapter)
        ;
...
snip
...
private RecipientListRouterSpec buildRecipientListRouterSpecForRules(RecipientListRouterSpec recipientListSpec,
        Collection<RuleMetadata> rules, MessageChannel publishedChannel) {

    // ??? How to subscribe this to publishedChannel??
    recipientListSpec
        .recipient(MessageChannels.publishSubscribe(this.getId().toString()+'.'+"mainReceiveChannel", stepTaskExecutor).get());
    
    rules.forEach(
            rule -> recipientListSpec.recipientFlow(getFilterExpression(rule), f -> createFlowDefForRule(f, rule)));
            
    recipientListSpec
        .ignoreSendFailures(true)
        .defaultOutputToParentFlow();
    
    return recipientListSpec;
}

Answer 1

publishedChannel必须作为输入通道传递给子流

        return flowDef
            .channel(receiveChannel) //  <---- This is the reference to the main publish channel in the child flow, which allows the builder to create the subscription
            .log()
            .handle(inboundAdapter)
... snip ...

            ;