Question

我正在做更多的RxJava实验，主要是试图找出适用于我的业务的设计模式。我创建了一个简单的航班跟踪应用程序，可跟踪多个航班，并在航班移动时作出相应的反应。

假设我有Collection<Flight>个Flight个对象。每个航班都有一个Observable<Point>，指定收到的最新坐标。如何从observable中提取最新的Flight对象本身，而不必将其保存到一个完整的单独变量中？ get()上没有Observable方法或类似方法吗？或者我的想法太强烈了？

public final class Flight {

    private final int flightNumber;
    private final String startLocation;
    private final String finishLocation;
    private final Observable<Point> observableLocation;
    private volatile Point currentLocation = new Point(0,0); //prefer not to have this

    public Flight(int flightNumber, String startLocation, String finishLocation) {

        this.flightNumber = flightNumber;
        this.startLocation = startLocation;
        this.finishLocation = finishLocation;
        this.observableLocation = FlightLocationManager.get().flightLocationFeed()
                .filter(f -> f.getFlightNumber() == this.flightNumber)
                .sample(1, TimeUnit.SECONDS)
                .map(f -> f.getPoint());

        this.observableLocation.subscribe(l -> currentLocation = l);
    }
    public int getFlightNumber() { 
        return flightNumber;
    }
    public String getStartLocation() { 
        return startLocation;
    }
    public String getFinishLocation() { 
        return finishLocation;
    }
    public Observable<Point> getObservableLocation() { 
        return observableLocation.last();
    }
    public Point getCurrentLocation() { 
        return currentLocation; //returns the latest observable location
        //would like to operate directly on observable instead of a cached value
    }
}

Answer 1

这是实现此目的的一种方法，主要是通过创建BlockingObservable。这是不稳定的，因为如果底层的observable没有完成，它有可能挂起：

observableLocation.toBlocking().last()

如何从Observable中提取最后一个值并将其返回？

1 个答案: