Question

我试图通过RabbitMQ获取rebus来检索我已成功发布到消息队列的消息。我不应该做什么，需要做什么。下面是示例代码？

static async Task MainSubscribeAsync()
        {

            var connection = "amqp://jhgj67546:yjyj5565@localhost";

            using (var activator = new BuiltinHandlerActivator())
            {
                activator.Register(() => new WagonHandler());

                var bus = Configure.With(activator)
                    .Logging(l => l.ColoredConsole())
                    .Transport(t => t.UseRabbitMq(connection, "wagon_v1")
                        .ExchangeNames(directExchangeName: "WamosExchange"))
                    .Start();

                await activator.Bus.Subscribe<Wagon>();

                Console.WriteLine("Done");
            }
        }

处理程序类看起来像这样

class WagonHandler : IHandleMessages
    {
        public async Task Handle(Wagon message)
        {
            Console.WriteLine($"Token {message.Token}");
            Console.WriteLine($"WagonId {message.WagonId}");
        }
    }

Answer 1

此代码

static async Task MainSubscribeAsync()
{

    var connection = "amqp://jhgj67546:yjyj5565@localhost";

    using (var activator = new BuiltinHandlerActivator())
    {
        activator.Register(() => new WagonHandler());

        var bus = Configure.With(activator)
            .Logging(l => l.ColoredConsole())
            .Transport(t => t.UseRabbitMq(connection, "wagon_v1")
                .ExchangeNames(directExchangeName: "WamosExchange"))
            .Start();

        await activator.Bus.Subscribe<Wagon>();

        Console.WriteLine("Done");
    }
}

启动总线，建立订阅，然后立即关闭，因为BuiltinHandlerActivator已被处理。

这意味着你可能很幸运，也许会收到一些消息，但在大多数情况下你可能会赢得<）>

您应该在应用程序启动时配置总线，然后在应用程序的整个生命周期内将其保留为单个实例（如the wiki page about Rebus' instance policies所述），仅在应用程序关闭时进行处理。

Rebus RabbitMQ无法订阅

1 个答案: