Spring amqp在侦听器外部拒绝消息

Question

当重新排队到原始队列时，消息可以再次返回到死信队列，并看到x-death报头计数不断增长。

由于某些原因，我们希望处理count>=5的死信html" target="_blank">消息（例如），并将其他消息重新排入死信队列。

我需要首先对消息进行基本的ack以检查X死亡计数头，然后将其发送到原始队列，如果计数足够大，否则在死信队列中重新排队。

我无法重新排队到死信队列，因为基本的get不在侦听器内部：抛出AmqpRejectAndDontRequeueException不起作用，因为异常不在rabbitmq侦听器对象内部抛出。

我尝试在receiveAndCallback方法中抛出异常,但这似乎并不更好：

rabbitTemplate.receiveAndReply(queueName, new ReceiveAndReplyCallback<Message, Object>() {

        @Override
        public Object handle(Message message) {
            Long messageXdeathCount = null;
            if (null != message.getMessageProperties() && null != message.getMessageProperties().getHeaders()) {
                List<Map<String, ?>> xdeathHeader =
                        (List<Map<String, ?>>) message.getMessageProperties().getHeaders().get(
                        "x-death");
                if (null != xdeathHeader && null != xdeathHeader.get(0)) {
                    messageXdeathCount = (Long) xdeathHeader.get(0).get("count");
                }
            }
            if (messageXdeathCount == null) {
                messageXdeathCount = 0L;
            }
            if (messageXdeathCount >= 5) {
                resendsMessage(message);
            } else {
                // this does not reject the message
                throw new AmqpRejectAndDontRequeueException("rejected");
            }
            return null;
        }
    });
    return receive;
}

在这个方法执行之后，消息没有像我预期的那样被拒绝，并且远离队列（它已经被加密）。

@Bean
public Exchange exchange() {
    TopicExchange exchange = new TopicExchange(EXCHANGE, true, false);
    admin().declareExchange(exchange);
    Map<String, Object> args = new HashMap<String, Object>();
    args.put("x-dead-letter-exchange", EXCHANGE);
    Queue queue = new Queue("queueName", true, false, false, args);
    admin().declareQueue(queue);
    Binding binding = BindingBuilder.bind(queue).to(exchange).with(routingKey).noargs();
    admin().declareBinding(binding);
    return exchange;
}

谢谢你的帮助

更新

我尝试了另一种方法，使用通道get和reject：

// exchange creation
@Bean
public Exchange exchange() throws IOException {
    Connection connection = connectionFactory().createConnection();
    Channel channel = channel();
    channel.exchangeDeclare(EXCHANGE, ExchangeTypes.TOPIC, true, false, null);
    Map<String, Object> args = new HashMap<String, Object>();
    args.put("x-dead-letter-exchange", EXCHANGE);
    channel.queueDeclare("queueName", true, false, false, args);
    channel.queueBind("queueName", EXCHANGE, routingKey);
    return exchange;
}

GetResponse response = channel.basicGet(queueName, false);
Long messageXdeathCount = null;
if(null != response.getProps() && null != response.getProps().getHeaders()) {
    List<Map<String, ?>> xdeathHeader =
            (List<Map<String, ?>>) response.getProps().getHeaders().get("x-death");
    if(null != xdeathHeader && null != xdeathHeader.get(0)) {
        messageXdeathCount = (Long) xdeathHeader.get(0).get("count");
    }
}
if (messageXdeathCount == null) {
    messageXdeathCount = 0L;
}
if (messageXdeathCount >= 5) {
    MessagePropertiesConverter messagePropertiesConverter = new DefaultMessagePropertiesConverter();
    MessageProperties messageProps =
            messagePropertiesConverter.toMessageProperties(response.getProps(),
response.getEnvelope(), "UTF-8");
    resendsMessage(new Message(response.getBody(), messageProps));
    channel.basicAck(response.getEnvelope().getDeliveryTag(), false);
} else {
    if(response.getProps().getHeaders().get("x-death") == null) {
        response.getProps().getHeaders().put("x-death", new ArrayList<>());
    }
    if(((List<Map<String, Object>>) response.getProps().getHeaders().get("x-death")).get(0) == null) {
        ((List<Map<String, Object>>)response.getProps().getHeaders().get("x-death")).add(new HashMap<>());
    }
    ((List<Map<String, Object>>) response.getProps().getHeaders().get("x-death")).get(0).put(
            "count", messageXdeathCount + 1);
    channel.basicReject(response.getEnvelope().getDeliveryTag(), true);
}

Answer 1

ReceiveAndReply()方法当前不提供对接收消息的确认的控制。请随意打开一个新的特性请求。

您可以改用侦听器容器来获得所需的灵活性。

编辑

rabbitTemplate.execute(channel -> {
    // basicGet, basicPublish, ack/nack etc here
});