Question

最近，我编写了一个演示程序，以结合Reactor和RabbitMQ启动反应式编程。这是我的演示代码：

public class FluxWithRabbitMQDemo {

private static final String QUEUE = "demo_thong";

private final reactor.rabbitmq.Sender sender;
private final Receiver receiver;

public FluxWithRabbitMQDemo() {
    this.sender = ReactorRabbitMq.createSender();
    this.receiver = ReactorRabbitMq.createReceiver();
}

public void run(int count) {
    ConnectionFactory connectionFactory = new ConnectionFactory();
    connectionFactory.useNio();
    SenderOptions senderOptions =  new SenderOptions()
            .connectionFactory(connectionFactory)
            .resourceCreationScheduler(Schedulers.elastic());

    reactor.rabbitmq.Sender sender = ReactorRabbitMq.createSender(senderOptions);

    Mono<AMQP.Queue.DeclareOk> queueDeclaration = sender.declareQueue(QueueSpecification.queue(QUEUE));
    Flux<Delivery> messages = receiver.consumeAutoAck(QUEUE);
    queueDeclaration.thenMany(messages).subscribe(m->System.out.println("Get message "+ new String(m.getBody())));



    Flux<OutboundMessageResult> dataStream = sender.sendWithPublishConfirms(Flux.range(1, count)
            .filter(m -> !m.equals(10))
            .parallel()
            .runOn(Schedulers.parallel())
            .doOnNext(i->System.out.println("Message  " + i + " run on thread "+Thread.currentThread().getId()))
            .map(i -> new OutboundMessage("", QUEUE, ("Message " + i).getBytes())));

    sender.declareQueue(QueueSpecification.queue(QUEUE))
            .thenMany(dataStream)
            .doOnError(e -> System.out.println("Send failed"+ e))
            .subscribe(m->{
                if (m!= null){
                    System.out.println("Sent successfully message "+new String(m.getOutboundMessage().getBody()));
                }
            });

    try {
        Thread.sleep(20000);
    } catch (InterruptedException e) {
        e.printStackTrace();
    }

}

public static void main(String[] args) throws Exception {
    int count = 20;
    FluxWithRabbitMQDemo sender = new FluxWithRabbitMQDemo();
    sender.run(count);
}

} 我希望在Flux发出项目后，发件人必须将其发送到RabbitMQ，并且在接收RabbitMQ之后，接收方必须接收它。但这一切都是顺序发生的，这就是我得到的结果

Message  3 run on thread 25
Message  4 run on thread 26
Message  8 run on thread 26
Message  13 run on thread 26
Message  17 run on thread 26
Message  2 run on thread 24
Message  1 run on thread 23
Message  6 run on thread 24
Message  5 run on thread 23
Message  9 run on thread 23
Message  14 run on thread 23
Message  18 run on thread 23
Message  11 run on thread 24
Message  15 run on thread 24
Message  19 run on thread 24
Message  7 run on thread 25
Message  12 run on thread 25
Message  16 run on thread 25
Message  20 run on thread 25
Sent successfully message Message 3
Sent successfully message Message 1
Sent successfully message Message 2
Sent successfully message Message 4
Sent successfully message Message 5
Sent successfully message Message 6
Sent successfully message Message 8
Sent successfully message Message 9
Sent successfully message Message 11
Sent successfully message Message 13
Sent successfully message Message 14
Sent successfully message Message 15
Sent successfully message Message 17
Sent successfully message Message 18
Sent successfully message Message 19
Sent successfully message Message 7
Sent successfully message Message 12
Sent successfully message Message 16
Sent successfully message Message 20
Get message Message 3
Get message Message 1
Get message Message 2
Get message Message 4
Get message Message 5
Get message Message 6
Get message Message 8
Get message Message 9
Get message Message 11
Get message Message 13
Get message Message 14
Get message Message 15
Get message Message 17
Get message Message 18
Get message Message 19
Get message Message 7
Get message Message 12
Get message Message 16
Get message Message 20

我不知道如何处理我的代码以达到预期的结果。有人能帮我吗？谢谢你的帮助！

Answer 1

消息生成太快。要查看交错，请在dataStream中添加

.doOnNext（i-> Thread.sleep（10））

使用Reactor和RabbitMQ进行反应式编程

1 个答案: