Question

我有一个JavaFX应用程序，它大量使用不同的媒体文件来显示信息。问题是，所有这些文件都包含在.zip存档中。如何访问这些文件并进行播放？这是我当前的MediaPlayerManager类。

public class MediaPlayerManager {

    private MediaPlayer mediaPlayer;
    private Media media;

    public void init(File file) {
        media = new Media(file.toURI().toString());
        mediaPlayer = new MediaPlayer(media);
        mediaPlayer.setAutoPlay(false);
        mediaView = new MediaView(mediaPlayer);
    }

    public void play() {
        mediaPlayer.play();
    }

    public void stop() {
        mediaPlayer.stop();
    }

    public void pause() {
        mediaPlayer.pause();
    }
}

这是我的Controller类中应该能够在存档中找到文件并播放它的方法。

public void playbackAudioFile(String audioPath, String audioName) {

    // audioPath is the path to archive which contains audioName

    MediaPlayerManager audioClip = new MediaPlayerManager();

    // Code missing for playing audio files inside .zip

    System.out.println("Playing sound...");
    audioClip.play();
    System.out.println("Sound played!");
}

Answer 1

借用Read directly a file within a Zip file - Java，您可以重构MediaPlayerManager，以便它直接接受URL的字符串形式：

public class MediaPlayerManager {

    private MediaPlayer mediaPlayer;
    private Media media;

    public void init(String url) {
        media = new Media(url);
        mediaPlayer = new MediaPlayer(media);
        mediaPlayer.setAutoPlay(false);
        mediaView = new MediaView(mediaPlayer);
    }

    public void init(File file) {
        init(file.toURI().toString());
    }

    // ...
}

现在做

public void playbackAudioFile(String audioPath, String audioName) {

    // audioPath is the path to archive which contains audioName

    MediaPlayerManager audioClip = new MediaPlayerManager();

    String zipURL = new File(audioPath).toURI().toString();
    String jarURL = "jar:" + zipURL + "!" +audioName ;

    audioClip.init(jarURL);

    System.out.println("Playing sound...");
    audioClip.play();
    System.out.println("Sound played!");
}

如何打开存档内的文件？

1 个答案: