Question

情境：

我向服务器队列发送请求（我使用RabbintMQ来异步处理消息）。当服务器处理请求时，它会对不同的队列产生两个响应。

我想使用RX订阅回复，但也可以访问相应的请求。

到目前为止我有什么：

我使用EventAggregator使用反应式扩展并为事件流公开IObservable：

public interface IEventAggregator
{
    void Publish<TEvent>(TEvent sampleEvent);
    IObservable<TEvent> GetEvent<TEvent>();
}

当我发送请求时，我发布了一个事件：

eventAggregator.Publish(new RequestSent { ID = Guid.NewGuid(), ... })
//I could later subscribe to the event like this:
//eventAggregator.GetEvent<RequestSent>().Subscribe(..)

当服务器响应时，响应也会发布到EventAggregator，因此我可以订阅它们：

eventAggregator.GetEvent<ResponseFromQueue1>().Subscribe(OnResponseFromQueue1)

eventAggregator.GetEvent<ResponseFromQueue2>().Subscribe(OnResponseFromQueue2)

我也可以订阅RequestSent

我需要什么：

private void OnResponseFromQueue1(RequestSent request, ResponseFromQueue1 response)
{
   I need access to both request and respone
}

这会更好：

private void OnResponse(
    RequestSent request, 
    ResponseFromQueue1 response1, 
    ResponseFromQueue2 response2)
{
   //actually, this would simplify implementation of my logic a lot
}

是否可以使用RX？

Answer 1

您可以使用SelectMany，假设您可以使用类似ID的内容将请求与回复相关联

trigger.SelectMany(requestData => {
  //We need to share this
  var id = Guid.NewGuid();

  //Publish your event
  eventAggregator.Publish(new Request { ID = id, /*...*/ });

  //Start listening for the two responses
  //Grab only the first item matching the IDs
  var left = eventAggregator.GetEvent<ResponseFromQueue1>().First(res => res.ID == id);
  var right = eventAggregator.GetEvent<ResponseFromQueue2>().First(res => res.Id == id);

  //We are done once both values have emitted.
  return left.Zip(right);
}, (request, responses) => {
  /*Here you will have access to the request and an array of the responses*/
});

要记住的一件事是，此代码现在假设Publish将在响应返回之前返回。既然你说这是RabbitMQ，这可能是一个安全的假设，但是如果你用它做任何单元测试就要记住了。

修改

在您的情景中，您实际上会看到：

//Set up the queue first eventAggregator.GetEvent<RequestSent>() .SelectMany(requestSent => { var id = requestSent.ID; var left = eventAggregator.GetEvent<ResponseFromQueue1>().First(res => res.ID == id); var right = eventAggregator.GetEvent<ResponseFromQueue2>().First(res => res.ID == id); return left.Zip(right); }, (request, response) => {/**/}); //...Sometime later eventAggregator.Publish(new Request{});

使用被动扩展来配置请求和响应

1 个答案: