与 TopShelf 作为 Windows 服务一起使用时，RabbitMQ 不接收消息

Question

我正在尝试将我的 RabbitMQ 微服务转换为 Windows 服务。我使用 TopShelf 进行转换。我的 RabbitMQ 微服务本身运行良好，但是当我将它作为服务运行时，它不再接收消息。在我的public static void Main(string[] args) 我有：

 HostFactory.Run(host =>
                {
                    host.Service<PersonService>(s =>                      
                    {
                        s.ConstructUsing(name => new PersonService());
                        s.WhenStarted(tc => tc.Start());             
                        s.WhenStopped(tc => tc.Stop());               
                    });
                    host.SetDescription("Windows service that provides database access totables."); 
                    host.SetDisplayName("Service");                   
                    host.SetServiceName("Service");
                });
            }

然后在我的PersonService 类中我有

public void Start() {
            ConsumeMessage();
        }

最后是我的ConsumeMessage 函数：

private static void ConsumeMessage() {
        MessagingConfig.SetInstance(new MessagingConstants());
        IMessageFactory pmfInst = MessageFactory.Instance;

        //message worker
        var factory = new ConnectionFactory() {
            HostName = MessagingConfig.Instance.GetBrokerHostName(),
            UserName = MessagingConfig.Instance.GetBrokerUserName(),
            Password = MessagingConfig.Instance.GetBrokerPassword()
        };

        var connection = factory.CreateConnection();

        using (var channel = connection.CreateModel()) {
            channel.QueueDeclare(queue: MessagingConfig.Instance.GetServiceQueueName(),
                                 durable: true,
                                 exclusive: false,
                                 autoDelete: false,
                                 arguments: null);

            channel.BasicQos(0, 1, false);

            var consumer = new EventingBasicConsumer(channel);

            channel.BasicConsume(queue: MessagingConfig.Instance.GetServiceQueueName(),
                                 noAck: false,
                                 consumer: consumer);

            Console.WriteLine("Service.");
            Console.WriteLine(" [x] Awaiting RPC requests");


            // Code Below Is Not Executed In Service
            consumer.Received += (model, ea) => {

                string response = null;

                var body = ea.Body;
                var props = ea.BasicProperties;
                var replyProps = channel.CreateBasicProperties();
                replyProps.CorrelationId = props.CorrelationId;

                string receivedMessage = null;

                try {
                    receivedMessage = Encoding.UTF8.GetString(body);
                    response = ProcessMessage(receivedMessage);
                }
                catch (Exception e) {
                    // Received message is not valid.
                    WinLogger.Log.Error(
                        "Errror Processing Message: " + receivedMessage + " :" + e.Message);

                    response = "";
                }
                finally {

                    var responseBytes = Encoding.UTF8.GetBytes(response);
                    channel.BasicPublish(exchange: "", routingKey: props.ReplyTo,
                    basicProperties: replyProps, body: responseBytes);
                    channel.BasicAck(deliveryTag: ea.DeliveryTag,
                    multiple: false);
                }
            };
            Console.ReadLine();
        }

查看A similar SO question，它似乎与 Windows 服务想要的返回有关，但我不确定如何调用 ConsumeMessage，因此执行了 consumer.Received += (model, ea) => {...};。

编辑：看起来我的阻塞机制Console.ReadLine(); 被服务忽略了，所以它只是继续并处理消息消费者。那么如何阻止那里接收消息呢？

Answer 1

您的代码使用using 构造，这意味着当您的OnStart 方法返回时，您的channel 实际上将被释放。 docs 建议在 OnStart 上进行初始化，因此在此处创建 channel 和 consumer，但不要使用 using：

this.connection = factory.CreateConnection();

this.channel = connection.CreateModel();
this.consumer = new EventingBasicConsumer(this.channel);

那么在OnStart 方法完成后这些对象将继续存在。您应该在 OnStop 方法中处理它们。