确定ActiveMQ异步消息失败？

我有ActiveMQ持久队列，由于性能原因，我将使用异步模式发布给生产者

 ActiveMQConnectionFactory factory = new ActiveMQConnectionFactory(brokerURL);
 factory.setUseAsyncSend(true);
 PooledConnectionFactory connectionFactory = new PooledConnectionFactory(factory);
 Connection connection = connectionFactory.createConnection();
 connection.start();
 Session session = connection.createSession(false, Session.AUTO_ACKNOWLEDGE);
 MessageProduer producer = session.createProducer(destination);
 Queue queue = session.createQueue(qName);
 producer.send(queue, message);

---

是否有一种方法可以注册处理程序以获得

producer.send（）

方法的错误/成功？

似乎jms 1.1规范不允许在send方法中注册回调，而jms 2.0允许它（）。由于ActiveMq基于jms 1.1，所以没有注册回调的标准方法。然而，ActvieMQ

javax.jms.MessageProducer

implementation

org.apache.activemq.ActiveMQMessageProducer

允许注册回调，我用它来创建我的解决方案（但不幸的是，我不得不放弃

PooledConnectionFactory

，因为无法从

PooledConnectionFactory

方法中获取

org.apache.activemq.ActiveMQMessageProducer

）

ActiveMQConnectionFactory factory = new ActiveMQConnectionFactory(brokerURL);
factory.setUseAsyncSend(true);
Connection connection = connectionFactory.createConnection();
connection.start();
Session session = connection.createSession(false, Session.AUTO_ACKNOWLEDGE);
MessageProduer producer = session.createProducer(destination);
Queue queue = session.createQueue(qName);
((ActiveMQMessageProducer)producer).send(queue, message, new AsyncCallback() {
        @Override
        public void onSuccess() {

        }

        @Override
        public void onException(JMSException exception) {

        }
    };);