将回调转换为被动发布者(Flux) [英] Convert a callback into a reactive publisher (Flux)

问题描述

我正在使用第三方库注册MessageListener，当某些事件发生时，它们调用已注册的监听器onMessage方法

public interface MessageListener {  
   // third party code, it auto-scans for all MessageListeners and registers them
    void onMessage(Message message);
}


public class SimpleMessageListener implements MessageListener {
   public void onMessage(Message message) {
      //do something non blocking
      //is it possible to 'transmit' to messagePublisher
}
   public Flux<Message> messagePublisher() {
       // a method to which to subscribeOn    
   }
}

所以我的问题是，将其转化为流量的最佳方法是什么

最后，我希望能够做这样的事情

messagePublisher().subscribe(System.out::println);

编辑* 我的第一次尝试是这样的

private List<FluxSink<Message>> handlers = new ArrayList<>();
public void onMessage(Message message) {
   handlers.forEach(han -> han.next(message));
}
public Flux<Message> messagePublisher() {
        return Flux.create(sink -> {
            handlers.add(sink);
            sink.onDispose(() -> handlers.remove(sink));
        });
    }

有效-但我觉得这不是一个很好的解决方案，让类实现FlosSink并手动处理是不是更好--目前我预计不会有很多订阅者。 但是有许多MessageListeners(每种类型一个)

推荐答案

您可以创建单个Flux实例来桥接MessageListener观察到的消息，例如

public class SimpleMessageListener implements MessageListener {
   private FluxSink<Message> handler;
   private Flux<Message> flux;

   public SimpleMessageListener() {
      flux = Flux.create(emitter -> {
          handler = emitter;
      }, OverflowStrategy.DROP); // or some other overflow strategy
   }

   public void onMessage(Message message) {
       if (handler != null) {
           /* 
            * null check is required to avoid NPE if a message is received 
            * before any subscription occurs since handler is instantiated
            * lazily when the first subscription is requested
            */
           handler.next(message);
       }
   }

   public Flux<Message> messagePublisher() {
       return flux;
   }
}

现在所有监听器都可以使用Flux‘publish()方法及其返回的ConnectableFlux订阅相同的messsagePublisher()Flux实例：

// fetch message publisher
Flux<Message> messagePublisher = messageListener.messagePublisher();

// prepare ConenctableFlux
ConnectableFlux<Message> connectableFlux = messagePublisher().publish();

// register subscribers
connectableFlux.subscribe(/* aConsumer */);
connectableFlux.subscribe(/* aCoreSubscriber */);
connectableFlux.subscribe(/* aSubscriber */);

// connect the ConnectableFlux to messagePublisher
connectableFlux.connect();

这篇关于将回调转换为被动发布者(Flux)的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！