使用java从jar作为资源播放.wav文件

Question

我想使用java代码播放.wav文件，该代码位于jar文件中作为资源。 我的代码看起来像这样 -

try {
     URL defaultSound = getClass().getResource("/images/ads/WindowsNavigationStart.wav");
     // getClass().getSy.getResource("/images/ads/WindowsNavigationStart.wav");
     File soundFile = new File(defaultSound.toURI());
     AudioInputStream audioInputStream = AudioSystem.getAudioInputStream(soundFile);
     Clip clip = AudioSystem.getClip();
     clip.open(audioInputStream);
     clip.start( );
} catch (Exception ex) {
     ex.printStackTrace();
}

WindowsNavigationStart.wav文件存在于我的一个jar文件中。 但得到以下例外 -

java.lang.IllegalArgumentException: URI is not hierarchical
at java.io.File.<init>(File.java:363)
at au.com.webscan.wwwizard.app.admin.customfile.UpOneLevelFolder.btnUpFolderActionPerformed(Unknown Source)
at au.com.webscan.wwwizard.app.admin.customfile.UpOneLevelFolder.access$000(Unknown Source)
at au.com.webscan.wwwizard.app.admin.customfile.UpOneLevelFolder$1.actionPerformed(Unknown Source)
at javax.swing.AbstractButton.fireActionPerformed(AbstractButton.java:1995)
at javax.swing.AbstractButton$Handler.actionPerformed(AbstractButton.java:2318)
at javax.swing.DefaultButtonModel.fireActionPerformed(DefaultButtonModel.java:387)
at javax.swing.DefaultButtonModel.setPressed(DefaultButtonModel.java:242)
at javax.swing.plaf.basic.BasicButtonListener.mouseReleased(BasicButtonListener.java:236)
at java.awt.AWTEventMulticaster.mouseReleased(AWTEventMulticaster.java:272)
at java.awt.Component.processMouseEvent(Component.java:6288)
at javax.swing.JComponent.processMouseEvent(JComponent.java:3267)
at java.awt.Component.processEvent(Component.java:6053)
at java.awt.Container.processEvent(Container.java:2041)
at java.awt.Component.dispatchEventImpl(Component.java:4651)
at java.awt.Container.dispatchEventImpl(Container.java:2099)
at java.awt.Component.dispatchEvent(Component.java:4481)
at java.awt.LightweightDispatcher.retargetMouseEvent(Container.java:4577)
at java.awt.LightweightDispatcher.processMouseEvent(Container.java:4238)
at java.awt.LightweightDispatcher.dispatchEvent(Container.java:4168)
at java.awt.Container.dispatchEventImpl(Container.java:2085)
at java.awt.Window.dispatchEventImpl(Window.java:2478)
at java.awt.Component.dispatchEvent(Component.java:4481)
at java.awt.EventQueue.dispatchEventImpl(EventQueue.java:643)
at java.awt.EventQueue.access$000(EventQueue.java:84)
at java.awt.EventQueue$1.run(EventQueue.java:602)
at java.awt.EventQueue$1.run(EventQueue.java:600)
at java.security.AccessController.doPrivileged(Native Method)
at java.security.AccessControlContext$1.doIntersectionPrivilege(AccessControlContext.java:87)
at java.security.AccessControlContext$1.doIntersectionPrivilege(AccessControlContext.java:98)
at java.awt.EventQueue$2.run(EventQueue.java:616)
at java.awt.EventQueue$2.run(EventQueue.java:614)
at java.security.AccessController.doPrivileged(Native Method)
at java.security.AccessControlContext$1.doIntersectionPrivilege(AccessControlContext.java:87)
at java.awt.EventQueue.dispatchEvent(EventQueue.java:613)
at java.awt.EventDispatchThread.pumpOneEventForFilters(EventDispatchThread.java:269)
at java.awt.EventDispatchThread.pumpEventsForFilter(EventDispatchThread.java:184)
at java.awt.EventDispatchThread.pumpEventsForHierarchy(EventDispatchThread.java:174)
at java.awt.EventDispatchThread.pumpEvents(EventDispatchThread.java:169)
at java.awt.EventDispatchThread.pumpEvents(EventDispatchThread.java:161)
at java.awt.EventDispatchThread.run(EventDispatchThread.java:122)

请给我一个解决方案。 谢谢你们。

Answer 1

你有没有尝试过：

InputStream is= getClass().getResourceAsStream("/images/ads/WindowsNavigationStart.wav");
AudioInputStream audioInputStream = AudioSystem.getAudioInputStream(is);

基本上我认为你不能在jar文件中用URI创建一个文件。 但是您可以直接传递输入流。

Answer 2

更改：

AudioInputStream audioInputStream = AudioSystem.getAudioInputStream(soundFile);

至：

System.out.println("defaultSound " + defaultSound);  // check the URL!
AudioInputStream audioInputStream = AudioSystem.getAudioInputStream(defaultSound);

Answer 3

使用Class.getResourceAsStream（）

获得inputStream的句柄后，获取audioInputStream并完成剩下的工作。

InputStream is = getClass().getResourceAsStream("......");
AudioInputStream ais = AudioSystem.getAudioInputStream(is);
Clip clip = AudioSystem.getClip();
clip.open(ais);

Answer 4

完美解决方案.......

URL url = this.getClass().getResource("sounds/beep.au");

String urls=url.toString(); 
urls=urls.replaceFirst("file:/", "file:///");

AudioClip ac=Applet.newAudioClip(new URL(urls));

ac.play();

Answer 5

在制作单jar java应用程序时，请参考我之前的回答。 标题是误导性的，但海报试图做一些几乎与你相同的事情。 一些最佳细节位于聊天记录的链接中。

Answer 6

     try {
        AudioPlayer.player.start(new AudioStream(getClass().getResourceAsStream("/sound/SystemNotification.wav")));
    } catch (Exception e) {
        e.printStackTrace();
    }

Answer 7

这对我来说很好：

public void playSound() {
        InputStream in;
        try {
            in = new BufferedInputStream(new FileInputStream(new File(
                    getClass().getClassLoader()
                            .getResource("com/kaito/resources/sound.wav").getPath())));
            AudioStream audioStream = new AudioStream(in);
            AudioPlayer.player.start(audioStream);
        } catch (FileNotFoundException e) {
            e.printStackTrace();
        } catch (IOException e) {
            e.printStackTrace();
        }
    }

Answer 8

以下允许我在Eclipse项目和导出的jar文件中播放声音：
- 注意使用BufferedInputStream
- 注意，使用inputStream而不是file。

在我的main（）中：

playAlarmSound();

在我班上：

public static void playAlarmSound() {
ClassLoader classLoader = App.class.getClassLoader();
InputStream inputStream = classLoader.getResourceAsStream("alarmsound.wav");
try {
  Clip clip = AudioSystem.getClip();
  AudioInputStream ais = AudioSystem.getAudioInputStream(new BufferedInputStream(inputStream));
  clip.open(ais);
  clip.start();
} catch (IOException | LineUnavailableException | UnsupportedAudioFileException e) {
  System.err.println("ERROR: Playing sound has failed");
  e.printStackTrace();
}
}

Answer 9

像Kal写道：

1. InputStream = getClass（）。getResourceAsStream（“......”）;
2. AudioInputStream ais = AudioSystem.getAudioInputStream（is）; ...

我这样做了，它起初没有用，但是“java.io.IOException”的问题是我使用了File.separator，由于某种原因，win 8.1无法处理“\\\\”......

public AudioInputStream getSound(String fileName){
    InputStream inputSound;
    AudioInputStream audioInStr;
    String fs = File.separator;

    try {
       absolutePath = fs +packageName+ fs +folderName+ fs +fileName;
       inputSound = getClass().getResourceAsStream(absolutePath);

       //if null pointer exception try unix, for some reason \\ doesn't work on win 8.1
       if(inputSound == null) {
           absolutePath = "/" + packageName + "/" + folderName + "/" + fileName;
           inputSound = getClass().getResourceAsStream(absolutePath);
       }

       audioInStr = AudioSystem.getAudioInputStream(new BufferedInputStream(inputSound));
       return audioInStr;
    }