Question

在下面的路线中，我希望来自queue1的10 msg应该同时处理，但是一次只能处理一个。

我怀疑错了吗？或做错了什么？

context.addRoutes(new RouteBuilder() {
        public void configure() {                                       
            from("test-jms:queue:test.queue1").threads(10)
            .process(sleep(1)); // sleep id is 1                
        }

        private Processor sleep(final int sleepId) {
            return new Processor() {                    
                @Override
                public void process(Exchange exchange) throws Exception {                       
                    System.out.println(curTime() + " Going for sleep sleepid=" + sleepId );
                    Thread.sleep(5000l);                        
                    System.out.println(curTime() + " Done sleep sleepid=" + sleepId );
                }
            };
        }

使用以下方式调用上述路线：

   ExecutorService ec = Executors.newFixedThreadPool(5);

    ec.submit(new Task(context,template));
    ec.submit(new Task(context,template));
    ec.submit(new Task(context,template));
    ec.submit(new Task(context,template));
    ec.submit(new Task(context,template));

static class Task  implements Runnable{
    CamelContext context;
    ProducerTemplate template;
    public Task(CamelContext context, ProducerTemplate template) {
        super();
        this.context = context;
        this.template = template;
    }
    @Override
    public void run() {         
           Exchange exchange = new DefaultExchange(context);
           exchange.setPattern(ExchangePattern.InOnly);
           exchange.getIn().setBody("Test Message: " + Thread.currentThread().getName());
           System.out.println(Thread.currentThread().getName());
           Exchange send = template.send("test-jms:queue:test.queue1",exchange);
           System.out.println("completed");           
    }

}

来自代码的OutPut：

10:24:11 Going for sleep sleepid=1
10:24:16 Done sleep sleepid=1

10:24:16 Going for sleep sleepid=1
10:24:21 Done sleep sleepid=1

10:24:21 Going for sleep sleepid=1
10:24:26 Done sleep sleepid=1

10:24:26 Going for sleep sleepid=1
10:24:31 Done sleep sleepid=1

10:24:31 Going for sleep sleepid=1
10:24:36 Done sleep sleepid=1

如果我们观察时间戳，我们将看到该路由仅在时间处理1 msg。

Answer 1

您需要在JMS端点上启用asyncConsumer以允许它为异步。执行此操作时，可以不按顺序处理从队列中消耗的消息，从而默认情况下订购消费者的原因。

代码应为

 public void configure() {                                       
            from("test-jms:queue:test.queue1?asyncConsumer=true").threads(10)
            .process(sleep(1)); // sleep id is 1                
        }

但是JMS组件具有内置并发性，通常可以更好地使用，因为它可以使用并发JMS使用者和并发网络。有关详细信息，请参阅选项concurrentConsumers和maxConcurrentConsumers。

在camel中使用队列进行线程DSL行为

1 个答案: