Question

请帮助我，我想用单个按钮使用yate fisher shuffle算法随机播放原始文件夹android studio的声音。我使用这段代码：

public void playSound() {
    MediaPlayer mp;
    int[] rawSoal = {R.raw.al, R.raw.fear, R.raw.love};
    shuffleArray(rawSoal);
    for (int i = 0; i < rawSoal.length; i++) ;
    try {

        Random random = new Random();
        mp = MediaPlayer.create(this, rawSoal[random.nextInt(rawSoal.length)]);
        mp.start();
        mp.release();

    }catch(Exception e){
        Log.e("ERROR", "Media Player", e);
        mp = null;
    }
}

实施Fisher-Yates shuffle

static void shuffleArray(int[] ar) {
    Random rnd = ThreadLocalRandom.current();
    for (int i = ar.length - 1; i > 0; i--) {
        int index = rnd.nextInt(i + 1);
        int a = ar[index];
        ar[index] = ar[i];
        ar[i] = a;
    }

}

通话按钮：

plybtn.setOnClickListener（new OnClickListener（）{

        @Override
        public void onClick(View arg0) {
            // TODO Auto-generated method stub
            playSound();
            mp.release();

        }
    });

Answer 1

试试这个：

public void playSound() {
    MediaPlayer mp;
    int[] rawSoal = {R.raw.al, R.raw.fear, R.raw.love};
    shuffleArray(rawSoal);
    try {
        int idx = new Random().nextInt(rawSoal.length);
        mp = MediaPlayer.create(this, rawSoal[idx]);
        mp.start();
        mp.release();

    }catch(Exception e){
        Log.e("ERROR", "Media Player", e);
        mp = null;
    }
}

static void shuffleArray(int[] arr)
    {
        Random rnd = new Random();
        for (int i = arr.length - 1; i > 0; i--)
        {
            int index = rnd.nextInt(i + 1);
            // Swap
            int a = arr[index];
            arr[index] = arr[i];
            arr[i] = a;
        }
    }

Answer 2

我认为因为你刚刚在start（）之后发布了MediaPlayer。所以音频没有播放。音频完成时释放MediaPlayer。

在原始文件夹中使用fisher yates shuffle算法在android上播放随机声音

2 个答案: