如何为ExoPlayer 2.x获取本地视频Uri

Question

我的文件res/raw夹中有一个dog.mp4视频文件，我想与ExoPlayer一起播放。我正在尝试从ExoPlayer开发人员指南（https://google.github.io/ExoPlayer/guide.html）中找出如何获取此代码行的视频Uri ：

MediaSource videoSource = new ExtractorMediaSource(mp4VideoUri,
    dataSourceFactory, extractorsFactory, null, null);

要获得它，我使用以下行：

Uri mp4VideoUri = Uri.parse("android.resources://"+getPackageName()+"/"+R.raw.dog);

还尝试了以下语法： android.resource://[package]/[res type]/[res name]

但是SimpleExoPlayerView保持黑色，我得到以下错误：

com.google.android.exoplayer2.upstream.HttpDataSource$HttpDataSourceException: Unable to connect to android.resources://lt.wilkas.deleteexoplayer/2131099648

我究竟做错了什么？

Answer 1

根据 ExoPlayer 2.12，它更容易使用MediaItem。为了创建它，我们需要Uri可以从RawResourceDataSource.buildRawResourceUri. 就是这样。只需设置此MediaItemdoprepare()并在准备好后即可播放：

SimpleExoPlayer.Builder(view.context).build().apply {
    val uri = RawResourceDataSource.buildRawResourceUri(R.raw.video)
    setMediaItem(MediaItem.fromUri(uri))
    prepare()
    playWhenReady = true
}

Answer 2

不要将您的视频从 raw 移动到任何其他文件夹

使用此代码播放视频：

    PlayerView playerView = findViewById(R.id.player_view);

    SimpleExoPlayer player = ExoPlayerFactory.newSimpleInstance(this);

    // Bind the player to the view.
    playerView.setPlayer(player);

    // Produces DataSource instances through which media data is loaded.
    DataSource.Factory dataSourceFactory = new DefaultDataSourceFactory(this, Util.getUserAgent(this, "yourApplicationName"));

    // This is the MediaSource representing the media to be played.
    MediaSource firstSource = new ExtractorMediaSource.Factory(dataSourceFactory).createMediaSource(RawResourceDataSource.buildRawResourceUri(R.raw.dog));

    // Prepare the player with the source.
    player.prepare(firstSource);

Answer 3

我发现该res/raw文件夹不能用于存储ExoPlayer的本地视频。它们应放在assets文件夹中。