我希望能够在我的程序中播放声音文件。我应该在哪里看？
答案 0 :(得分：120)
我编写了以下代码，工作正常。但我认为它只适用于.wav格式。
public static synchronized void playSound(final String url) {
new Thread(new Runnable() {
// The wrapper thread is unnecessary, unless it blocks on the
// Clip finishing; see comments.
public void run() {
try {
Clip clip = AudioSystem.getClip();
AudioInputStream inputStream = AudioSystem.getAudioInputStream(
Main.class.getResourceAsStream("/path/to/sounds/" + url));
clip.open(inputStream);
clip.start();
} catch (Exception e) {
System.err.println(e.getMessage());
}
}
}).start();
}
答案 1 :(得分：24)
Java Tutorial的Sound Trail值得作为起点。
答案 2 :(得分：20)
一个不好的例子：
import sun.audio.*; //import the sun.audio package
import java.io.*;
//** add this into your application code as appropriate
// Open an input stream to the audio file.
InputStream in = new FileInputStream(Filename);
// Create an AudioStream object from the input stream.
AudioStream as = new AudioStream(in);
// Use the static class member "player" from class AudioPlayer to play
// clip.
AudioPlayer.player.start(as);
// Similarly, to stop the audio.
AudioPlayer.player.stop(as);
答案 3 :(得分：9)
我不想拥有这么多行代码只是为了播放一个简单的该死的声音。如果你有JavaFX包（已经包含在我的jdk 8中），这可以工作。
private static void playSound(String sound){
// cl is the ClassLoader for the current class, ie. CurrentClass.class.getClassLoader();
URL file = cl.getResource(sound);
final Media media = new Media(file.toString());
final MediaPlayer mediaPlayer = new MediaPlayer(media);
mediaPlayer.play();
}
注意：您需要initialize JavaFX。一个快速的方法是在你的应用程序中调用一次JFXPanel（）的构造函数：
static{
JFXPanel fxPanel = new JFXPanel();
}
答案 4 :(得分：7)
要在java中播放声音，可以参考以下代码。
import java.io.*;
import java.net.URL;
import javax.sound.sampled.*;
import javax.swing.*;
// To play sound using Clip, the process need to be alive.
// Hence, we use a Swing application.
public class SoundClipTest extends JFrame {
public SoundClipTest() {
this.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
this.setTitle("Test Sound Clip");
this.setSize(300, 200);
this.setVisible(true);
try {
// Open an audio input stream.
URL url = this.getClass().getClassLoader().getResource("gameover.wav");
AudioInputStream audioIn = AudioSystem.getAudioInputStream(url);
// Get a sound clip resource.
Clip clip = AudioSystem.getClip();
// Open audio clip and load samples from the audio input stream.
clip.open(audioIn);
clip.start();
} catch (UnsupportedAudioFileException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
} catch (LineUnavailableException e) {
e.printStackTrace();
}
}
public static void main(String[] args) {
new SoundClipTest();
}
}
答案 5 :(得分：4)
无论出于何种原因，当我调用this.getClass（）。getResourceAsStream（）时，wchargin的最佳答案是给我一个空指针错误。
对我有用的是：
void playSound(String soundFile) {
File f = new File("./" + soundFile);
audioIn = AudioSystem.getAudioInputStream(f.toURI().toURL());
Clip clip = AudioSystem.getClip();
clip.open(audioIn);
clip.start();
}
我会播放声音：
playSound("sounds/effects/sheep1.wav");
sounds / effects / sheep1.wav位于Eclipse项目的基本目录中（因此不在src文件夹中）。
答案 6 :(得分：3)
可以选择导入适用于applet和应用程序的声音文件：将音频文件转换为.java文件，只需在代码中使用它们。
我开发了一种工具，使这个过程变得更加容易。它简化了Java Sound API。
答案 7 :(得分：3)
我之前创建了一个游戏框架，可以在Android和桌面上工作，处理声音的桌面部分也可以作为你需要的灵感。
以下是供参考的代码。
package com.athanazio.jaga.desktop.sound;
import java.io.BufferedInputStream;
import java.io.IOException;
import java.io.InputStream;
import javax.sound.sampled.AudioFormat;
import javax.sound.sampled.AudioInputStream;
import javax.sound.sampled.AudioSystem;
import javax.sound.sampled.DataLine;
import javax.sound.sampled.LineUnavailableException;
import javax.sound.sampled.SourceDataLine;
import javax.sound.sampled.UnsupportedAudioFileException;
public class Sound {
AudioInputStream in;
AudioFormat decodedFormat;
AudioInputStream din;
AudioFormat baseFormat;
SourceDataLine line;
private boolean loop;
private BufferedInputStream stream;
// private ByteArrayInputStream stream;
/**
* recreate the stream
*
*/
public void reset() {
try {
stream.reset();
in = AudioSystem.getAudioInputStream(stream);
din = AudioSystem.getAudioInputStream(decodedFormat, in);
line = getLine(decodedFormat);
} catch (Exception e) {
e.printStackTrace();
}
}
public void close() {
try {
line.close();
din.close();
in.close();
} catch (IOException e) {
}
}
Sound(String filename, boolean loop) {
this(filename);
this.loop = loop;
}
Sound(String filename) {
this.loop = false;
try {
InputStream raw = Object.class.getResourceAsStream(filename);
stream = new BufferedInputStream(raw);
// ByteArrayOutputStream out = new ByteArrayOutputStream();
// byte[] buffer = new byte[1024];
// int read = raw.read(buffer);
// while( read > 0 ) {
// out.write(buffer, 0, read);
// read = raw.read(buffer);
// }
// stream = new ByteArrayInputStream(out.toByteArray());
in = AudioSystem.getAudioInputStream(stream);
din = null;
if (in != null) {
baseFormat = in.getFormat();
decodedFormat = new AudioFormat(
AudioFormat.Encoding.PCM_SIGNED, baseFormat
.getSampleRate(), 16, baseFormat.getChannels(),
baseFormat.getChannels() * 2, baseFormat
.getSampleRate(), false);
din = AudioSystem.getAudioInputStream(decodedFormat, in);
line = getLine(decodedFormat);
}
} catch (UnsupportedAudioFileException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
} catch (LineUnavailableException e) {
e.printStackTrace();
}
}
private SourceDataLine getLine(AudioFormat audioFormat)
throws LineUnavailableException {
SourceDataLine res = null;
DataLine.Info info = new DataLine.Info(SourceDataLine.class,
audioFormat);
res = (SourceDataLine) AudioSystem.getLine(info);
res.open(audioFormat);
return res;
}
public void play() {
try {
boolean firstTime = true;
while (firstTime || loop) {
firstTime = false;
byte[] data = new byte[4096];
if (line != null) {
line.start();
int nBytesRead = 0;
while (nBytesRead != -1) {
nBytesRead = din.read(data, 0, data.length);
if (nBytesRead != -1)
line.write(data, 0, nBytesRead);
}
line.drain();
line.stop();
line.close();
reset();
}
}
} catch (IOException e) {
e.printStackTrace();
}
}
}
答案 8 :(得分：1)
我很惊讶没有人建议使用Applet。 Use Applet。您必须将蜂鸣声音频文件作为wav文件提供，但是它可以工作。我在Ubuntu上尝试过：
package javaapplication2;
import java.applet.Applet;
import java.applet.AudioClip;
import java.io.File;
import java.net.MalformedURLException;
import java.net.URL;
public class JavaApplication2 {
public static void main(String[] args) throws MalformedURLException {
File file = new File("/path/to/your/sounds/beep3.wav");
URL url = null;
if (file.canRead()) {url = file.toURI().toURL();}
System.out.println(url);
AudioClip clip = Applet.newAudioClip(url);
clip.play();
System.out.println("should've played by now");
}
}
//beep3.wav was available from: http://www.pacdv.com/sounds/interface_sound_effects/beep-3.wav
答案 9 :(得分：0)
对我有用。简单变体
public void makeSound(){
File lol = new File("somesound.wav");
try{
Clip clip = AudioSystem.getClip();
clip.open(AudioSystem.getAudioInputStream(lol));
clip.start();
} catch (Exception e){
e.printStackTrace();
}
}
答案 10 :(得分：0)
import java.net.URL;
import java.net.MalformedURLException;
import javax.sound.sampled.AudioInputStream;
import javax.sound.sampled.AudioSystem;
import javax.sound.sampled.AudioFormat;
import javax.sound.sampled.Clip;
import javax.sound.sampled.LineUnavailableException;
import javax.sound.sampled.UnsupportedAudioFileException;
import java.io.IOException;
import java.io.File;
public class SoundClipTest{
//plays the sound
public static void playSound(final String path){
try{
final File audioFile=new File(path);
AudioInputStream audioIn=AudioSystem.getAudioInputStream(audioFile);
Clip clip=AudioSystem.getClip();
clip.open(audioIn);
clip.start();
long duration=getDurationInSec(audioIn);
//System.out.println(duration);
//We need to delay it otherwise function will return
//duration is in seconds we are converting it to milliseconds
Thread.sleep(duration*1000);
}catch(LineUnavailableException | UnsupportedAudioFileException | MalformedURLException | InterruptedException exception){
exception.printStackTrace();
}
catch(IOException ioException){
ioException.printStackTrace();
}
}
//Gives duration in seconds for audio files
public static long getDurationInSec(final AudioInputStream audioIn){
final AudioFormat format=audioIn.getFormat();
double frameRate=format.getFrameRate();
return (long)(audioIn.getFrameLength()/frameRate);
}
////////main//////
public static void main(String $[]){
//SoundClipTest test=new SoundClipTest();
SoundClipTest.playSound("/home/dev/Downloads/mixkit-sad-game-over-trombone-471.wav");
}
}
答案 11 :(得分：-1)
这个帖子比较陈旧，但我确定了一个可能有用的选项。
您可以使用外部程序（如Windows Media Player或VLC），而不是使用Java AudioStream库，并通过Java使用控制台命令运行它。
String command = "\"C:/Program Files (x86)/Windows Media Player/wmplayer.exe\" \"C:/song.mp3\"";
try {
Process p = Runtime.getRuntime().exec(command);
catch (IOException e) {
e.printStackTrace();
}
这也将创建一个可以控制程序的单独进程。
p.destroy();
当然，这比使用内部库需要更长的时间，但是可能有一些程序可以更快地启动，并且可能没有给定某些控制台命令的GUI。
如果时间不重要，那么这很有用。