消息队列Pulsar入门(三) 探究不同消费模式下的ACK失败重试机制,附源码演示

最新推荐文章于 2025-10-26 15:10:35 发布

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最新推荐文章于 2025-10-26 15:10:35 发布 · 1.5k 阅读

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#java #jvm #服务器

单机部署

先看一下最简单的单机情况下的例子,对ACK失败重试这一套回顾下
consumer.acknowledge(msg);  //消费成功确认
consumer.negativeAcknowledge(msg);  //消费失败确认
模拟单机出现消费失败的情况 int i=1/0,会看到后续消息会再次被投递消费
复制代码

@Test
public void testProduce() throws Exception {
    Producer<String> producer = client.newProducer(Schema.STRING)
            .topic("my-topic")
            .enableBatching(false)
            .create();
    producer.send("my-topic发送消息" + System.currentTimeMillis());
    System.in.read();
}

/**
 * 单机模式下的失败重试
 */
@Test
public void testConsumerDanJI() throws IOException {
    MessageListener myMessageListener = (consumer, msg) -> {
        try {
            System.out.println("Message received my-subscription1 : " + new String(msg.getData()));
            //模拟消费失败的情况
            int i =1/0;
            consumer.acknowledge(msg);
        } catch (Exception e) {
            consumer.negativeAcknowledge(msg);
        }
    };
    Consumer consumer = client.newConsumer()
            .topic("my-topic")
            .subscriptionName("my-subscription1")
            .messageListener(myMessageListener)
            .subscribe();
    System.in.read();
}
复制代码

Exclusive

Exclusive模式例如有集群A,B两台机器,正常A,B都会收到消息,现在模拟A消费失败
看下后续是A,B两台机器都会重新消费 还是只有A会??
答案是只有A会
复制代码

/**
 * Exclusive下的失败重试
 */
@Test
public void testConsumerExclusive() throws IOException {
    MessageListener myMessageListener1 = (consumer, msg) -> {
        try {
            System.out.println("Message received my-subscription1 : " + new String(msg.getData()));
            //模拟消费失败的情况
            int i=1/0;
            consumer.acknowledge(msg);
        } catch (Exception e) {
            consumer.negativeAcknowledge(msg);
        }
    };
    Consumer consumer = client.newConsumer()
            .topic("my-topic")
            .subscriptionName("my-subscription1")
            .subscriptionType(SubscriptionType.Exclusive)
            .messageListener(myMessageListener1)
            .subscribe();

    Me

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