For the latest stable version, please use Spring Integration 6.4.1!

Sub-flows support

Some of if…​else and publish-subscribe components provide the ability to specify their logic or mapping by using sub-flows. The simplest sample is .publishSubscribeChannel(), as the following example shows:

@Bean
public IntegrationFlow subscribersFlow() {
    return flow -> flow
            .publishSubscribeChannel(Executors.newCachedThreadPool(), s -> s
                    .subscribe(f -> f
                            .<Integer>handle((p, h) -> p / 2)
                            .channel(c -> c.queue("subscriber1Results")))
                    .subscribe(f -> f
                            .<Integer>handle((p, h) -> p * 2)
                            .channel(c -> c.queue("subscriber2Results"))))
            .<Integer>handle((p, h) -> p * 3)
            .channel(c -> c.queue("subscriber3Results"));
}

You can achieve the same result with separate IntegrationFlow @Bean definitions, but we hope you find the sub-flow style of logic composition useful. We find that it results in shorter (and so more readable) code.

Starting with version 5.3, a BroadcastCapableChannel-based publishSubscribeChannel() implementation is provided to configure sub-flow subscribers on broker-backed message channels. For example, we now can configure several subscribers as sub-flows on the Jms.publishSubscribeChannel():

@Bean
public JmsPublishSubscribeMessageChannelSpec jmsPublishSubscribeChannel() {
    return Jms.publishSubscribeChannel(jmsConnectionFactory())
                .destination("pubsub");
}

@Bean
public IntegrationFlow pubSubFlow(BroadcastCapableChannel jmsPublishSubscribeChannel) {
    return f -> f
            .publishSubscribeChannel(jmsPublishSubscribeChannel,
                    pubsub -> pubsub
                            .subscribe(subFlow -> subFlow
                                .channel(c -> c.queue("jmsPubSubBridgeChannel1")))
                            .subscribe(subFlow -> subFlow
                                .channel(c -> c.queue("jmsPubSubBridgeChannel2"))));
}

A similar publish-subscribe sub-flow composition provides the .routeToRecipients() method.

Another example is using .discardFlow() instead of .discardChannel() on the .filter() method.

The .route() deserves special attention. Consider the following example:

@Bean
public IntegrationFlow routeFlow() {
    return f -> f
            .<Integer, Boolean>route(p -> p % 2 == 0,
                    m -> m.channelMapping("true", "evenChannel")
                            .subFlowMapping("false", sf ->
                                    sf.<Integer>handle((p, h) -> p * 3)))
            .transform(Object::toString)
            .channel(c -> c.queue("oddChannel"));
}

The .channelMapping() continues to work as it does in regular Router mapping, but the .subFlowMapping() tied that sub-flow to the main flow. In other words, any router’s sub-flow returns to the main flow after .route().

Sometimes, you need to refer to an existing IntegrationFlow @Bean from the .subFlowMapping(). The following example shows how to do so:

@Bean
public IntegrationFlow splitRouteAggregate() {
    return f -> f
            .split()
            .<Integer, Boolean>route(o -> o % 2 == 0,
                    m -> m
                            .subFlowMapping(true, oddFlow())
                            .subFlowMapping(false, sf -> sf.gateway(evenFlow())))
            .aggregate();
}

@Bean
public IntegrationFlow oddFlow() {
    return f -> f.handle(m -> System.out.println("odd"));
}

@Bean
public IntegrationFlow evenFlow() {
    return f -> f.handle((p, h) -> "even");
}


In this case, when you need to receive a reply from such a sub-flow and continue the main flow, this IntegrationFlow bean reference (or its input channel) has to be wrapped with a .gateway() as shown in the preceding example. The oddFlow() reference in the preceding example is not wrapped to the .gateway(). Therefore, we do not expect a reply from this routing branch. Otherwise, you end up with an exception similar to the following:

Caused by: org.springframework.beans.factory.BeanCreationException:
    The 'currentComponent' (org.springframework.integration.router.MethodInvokingRouter@7965a51c)
    is a one-way 'MessageHandler' and it isn't appropriate to configure 'outputChannel'.
    This is the end of the integration flow.

When you configure a sub-flow as a lambda, the framework handles the request-reply interaction with the sub-flow and a gateway is not needed.

Sub-flows can be nested to any depth, but we do not recommend doing so. In fact, even in the router case, adding complex sub-flows within a flow would quickly begin to look like a plate of spaghetti and be difficult for a human to parse.

In cases where the DSL supports a subflow configuration, when a channel is normally needed for the component being configured, and that subflow starts with a channel() element, the framework implicitly places a bridge() between the component output channel and the flow’s input channel. For example, in this filter definition:

.filter(p -> p instanceof String, e -> e
	.discardFlow(df -> df
                         .channel(MessageChannels.queue())
                         ...)

the Framework internally creates a DirectChannel bean for injecting into the MessageFilter.discardChannel. Then it wraps the subflow into an IntegrationFlow starting with this implicit channel for the subscription and places a bridge before the channel() specified in the flow. When an existing IntegrationFlow bean is used as a subflow reference (instead of an inline subflow, e.g. a lambda), there is no such bridge required because the framework can resolve the first channel from the flow bean. With an inline subflow, the input channel is not yet available.