Question

我有一个Sound类，其中包含一个媒体播放器，我想编写一个接收声音列表并全部播放的函数，该函数应返回一个可完成

interface MediaPlayer {
    fun play(): Completable
}

class Sound(val title, val mediaPlayer: MediaPlayer)

//In other class, we have a list of sound to play
val soundList = List<Sound>(mockSound1, mockSound2,..,mockSound10)

fun playSound(): Completable {
    return mockSound1.play()
}

fun playAllSounds(): Completable {
    soundList.forEach(sound -> sound.mediaPlayer.play()) //Each of this will return Completable. 

//HOW to return Completable
return ??? do we have somthing like zip(listOf<Completable>)
}


//USE
playSound().subscrible(...) //Works well

playAllSounds().subscribe()???

Answer 1

您可以从文档中使用concat

返回一个Completable，该Completable仅在所有源都完成后才完成。

您可以执行以下操作：

fun playAllSounds(): Completable {
    val soundsCompletables = soundList.map(sound -> sound.mediaPlayer.play())
    return Completable.concat(soundCompletables)
}

参考：http://reactivex.io/RxJava/javadoc/io/reactivex/Completable.html#concat-java.lang.Iterable-

Answer 2

您可以尝试Completable.merge。它将立即订阅所有完成项。这是link to the docs

邮编列表可填写

2 个答案: