EasyNetQ（RabbitMQ）在处理消息时，如果抛出异常，继续不断发送到订阅队列，不断处理（也就是不自动确认消息已到达）

默认情况下，EasyNetQ的消息处理过程中，如果throw exception，那么，依然是认为消息已经送达，不会再次推送，为了让RabbitMQ再次推送，可以这么实现：

        public sealed class AlwaysRequeueErrorStrategy : IConsumerErrorStrategy
        {
            public void Dispose()
            {
            }

            public AckStrategy HandleConsumerError(ConsumerExecutionContext context, Exception exception)
            {
                return AckStrategies.NackWithRequeue;
            }

            public AckStrategy HandleConsumerCancelled(ConsumerExecutionContext context)
            {
                return AckStrategies.NackWithRequeue;
            }
        }

        [TestMethod]
        public void test()
        {
            var hostName = _configuration["RabbitMQ:HostName"];
            var mqport = _configuration["RabbitMQ:Port"];
            var userName = _configuration["RabbitMQ:UserName"];
            var password = _configuration["RabbitMQ:Password"];

            var connectionConfiguration = new ConnectionConfiguration
            {

                Hosts = new List<HostConfiguration>
                {
                    new HostConfiguration
                    {
                        Host = hostName,
                        Port = Convert.ToUInt16(mqport)
                    }
                },
                Port = Convert.ToUInt16(mqport),
                VirtualHost = "/",
                UserName = userName,
                Password = password
            };
            var _bus = RabbitHutch.CreateBus(connectionConfiguration , x=> {
                x.Register<IConsumerErrorStrategy, AlwaysRequeueErrorStrategy>();
            });

            var exchange = _bus.Advanced.ExchangeDeclare("ExchangeNAME_test", ExchangeType.Topic);
            
            var nickNameQu = _bus.Advanced.QueueDeclare("queuename2222");
            int count = 0;
            _bus.Advanced.Bind(exchange, nickNameQu, "RouterKey1");
            _bus.Advanced.Consume(nickNameQu, (data, properties, info) =>
            {
                count++;
                if (count < 2)
                    throw new Exception("my error");
                 var msg = Encoding.UTF8.GetString(data);
            });

            Task.Run(() => {
                _bus.PublishAsync(new UpdateAvaterMessage() {  Avater = "abcc22" }, "RouterKey1");
            });
           
            Thread.Sleep(100000);
        }

原理是重写IConsumerErrorStrategy

（如果不是使用EasyNetQ，传统的RabbitMQ库是用手动ack实现）

另外，如果消息处理过程中，服务器死机，这种情况消息是会重发的，不需要担心