如何打开档案内的文件？

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压缩文件中。我如何访问这些文件并播放它们？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(); 
    } 
} 

这是我的控制器类应该能够找到文件内的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，可以重构MediaPlayerManager，使其接受URL直接的字符串形式：

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!"); 
}