Question

我有一个不那么漂亮的RxJava流程： - （

RxView.clicks(sendButton)
      .flatMap(ignored -> cameraManager.getCamera())
      .flatMap(camera -> camera.isReachable())
      .filter(reachable -> reachable)
      .flatMap(ignored -> cameraManager.getCamera())
      .flatMap(camera -> camera.sendCommand("foo bar"))
      .subscribe(send -> Timber.i("send successful"),
                 error -> Timber.e(error, "send failed: "));

我不喜欢cameraManager.getCamera()的这个链对镜头做些什么。

flatMap是否有办法让flatMap(camera -> Pair.create(camera, camera.isReachable()))同时返回项目及其转化结果？

RxView.clicks(sendButton)
      .flatMap(ignored -> cameraManager.getCamera())
      .flatMap(camera -> Pair.create(camera, camera.isReachable()))
      .filter(pair -> pair.second)
      .flatMap(pair -> pair.first.sendCommand("foo bar"))
      .subscribe(send -> Timber.i("send successful"),
                 error -> Timber.e(error, "send failed: "));

了解我的代码段的更多代码：

private class Bar {

    private CameraManager cameraManager = new CameraManager();
    private View          sendButton    = null;

    public Bar() {
        bind();
    }

    private void bind() {

        RxView.clicks(sendButton)
              .flatMap(ignored -> cameraManager.getCamera())
              .flatMap(camera -> camera.isReachable())
              .filter(reachable -> reachable)
              .flatMap(ignored -> cameraManager.getCamera())
              .flatMap(camera -> camera.sendCommand("foo bar"))
              .subscribe(send -> Timber.i("send successful"),
                         error -> Timber.e(error, "send failed: "));
    }
}

private class CameraManager {

    public Observable<Camera> getCamera() {
        return Observable.just(new Camera());
    }
}

public class Camera {

    private BehaviorSubject<Boolean> reachable = BehaviorSubject.create(false);


    public Observable<Boolean> isReachable() {
        return reachable.asObservable();
    }

    public Observable<Boolean> sendCommand(String msg) {
        return Observable.create(subscriber -> {

            // send message...
            final boolean sendSuccessful = true;

            if (sendSuccessful) {
                subscriber.onNext(true);
                subscriber.onCompleted();
            } else {
                subscriber.onError(new Exception());
            }
        });
    }
}

Answer 1

有一个flatMap重载，它有一个额外的Func2<T, U, R>，它接收原始元素和生成的Observable<U>的每个元素，并返回一个要发出的值。

Observable.range(1, 10)
.flatMap(v -> Observable.range(1, 2), (v, w) -> v + 100 * w)
.subscribe(System.out.println);

flatMap返回项目和转换结果

1 个答案: