Question

我正在尝试从文件路径创建媒体播放器

private void play(String path) {
    MediaPlayer player = null;
    if (path.startsWith("assets/music")) {
        path = path.replace("assets/", "");
        AssetFileDescriptor afd;
        try {
            afd = mActivity.getAssets().openFd(path);
            player = new MediaPlayer();
            player.setDataSource(afd.getFileDescriptor(),
                    afd.getStartOffset(), afd.getLength());
            player.prepare();
        } catch (IOException e) {
            Log.e(TAG, e.toString());
        }

    } else {

        player = MediaPlayer.create(mActivity, Uri.parse(path));


    }
    if (player != null) {
        player.setOnCompletionListener(releaseListener);
        player.setOnErrorListener(new OnErrorListener() {

            @Override
            public boolean onError(MediaPlayer mp, int what, int extra) {
                Log.e(TAG, "mp error: " + what + ", extra: " + extra);
                return true;
            }
        });
        player.start();
        mPlayer = player;
    }
}

从资产路径

它的工作原理但是MediaPlayer.create（mActivity，Uri.parse（path））;总是返回null

完整路径

/ storage / emulated / 0 /下载/ Buzzer - 1.mp3

资产路径

assets / music / Buzzer - 1.mp3

Answer 1

如果您说传递给path的{{1}}的值为Uri.parse(path)，则该值不是有效/storage/emulated/0/Download/Buzzer - 1.mp3。值得注意的是，没有计划。之一：

传递指向所需文件的Uri并使用File或
确保Uri.fromFile()有一个方案（例如path）

MediaPlayer.create（）为filepath返回null

1 个答案: