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Zip列表可完成

rx-java android

coinhndp · 技术社区 · 7 年前

我有一个声音类,它包含一个媒体播放器,我想写一个函数,接收声音列表并播放它们,该函数应该返回一个完全的

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()???

1 回复 | 直到 7 年前

Ahmed Ashraf 7 年前

你可以用 concat

返回一个Completable,它仅在所有源相继完成时完成。

你可以这样做:

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

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