android - zip 列表完成

Question

我有一个包含媒体播放器的 Sound 类，我想编写一个函数来接收声音列表并全部播放，该函数应该返回一个可完成

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

您可以使用 concat，来自文档

Returns a Completable which completes only when all sources complete, one after another.

你可以这样做:

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

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