Java 如何读取音频文件？我应该使用哪种方法？
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How to read an audio file? Which method should I use?
提问by Tofuw
I have a panel with 2 buttons. When I click on the button 1, I'd simply like to read an audio file (a .WAV in that case). Then, when I click on the button 2, I'd like to stop the music.
我有一个带有 2 个按钮的面板。当我单击按钮 1 时，我只想读取音频文件（在这种情况下为 .WAV）。然后，当我点击按钮 2 时，我想停止播放音乐。
I do some research, but I'm a little confused about the different methods. Which one is the best in my case ? Can someone explains the difference between AudioClip, JavaSound and JavaMediaFramework please ?
我做了一些研究，但我对不同的方法有点困惑。在我的情况下，哪一个最好？有人可以解释一下 AudioClip、JavaSound 和 JavaMediaFramework 之间的区别吗？
I've also try an example, but it contains errors.
我也试过一个例子，但它包含错误。
Here is my Main.class:
这是我的Main.class：
import java.io.ByteArrayInputStream;
import java.io.InputStream;
public class Main
{
public static void main(String[] args)
{
SoundPlayer player = new SoundPlayer("C:/Documents and Settings/All Users/Documents/Ma musique/échantillons de musique/Symphonie n° 9 de Beethoven (scherzo).wma");
InputStream stream = new ByteArrayInputStream(player.getSamples());
player.play(stream);
}
}
Here is my SoundPlayer.class:
这是我的SoundPlayer.class：
import java.io.DataInputStream;
import java.io.File;
import java.io.IOException;
import java.io.InputStream;
import javax.sound.sampled.*;
public class SoundPlayer
{
private AudioFormat format;
private byte[] samples;
/**
*
* @param filename le lien vers le fichier song (URL ou absolute path)
*/
public SoundPlayer(String filename)
{
try
{
AudioInputStream stream = AudioSystem.getAudioInputStream(new File(filename));
format = stream.getFormat();
samples = getSamples(stream);
}
catch (UnsupportedAudioFileException e){e.printStackTrace();}
catch (IOException e){e.printStackTrace();}
}
public byte[] getSamples()
{
return samples;
}
public byte[] getSamples(AudioInputStream stream)
{
int length = (int)(stream.getFrameLength() * format.getFrameSize());
byte[] samples = new byte[length];
DataInputStream in = new DataInputStream(stream);
try
{
in.readFully(samples);
}
catch (IOException e){e.printStackTrace();}
return samples;
}
public void play(InputStream source)
{
int bufferSize = format.getFrameSize() * Math.round(format.getSampleRate() / 10);
byte[] buffer = new byte[bufferSize];
SourceDataLine line;
try
{
DataLine.Info info = new DataLine.Info(SourceDataLine.class, format);
line = (SourceDataLine)AudioSystem.getLine(info);
line.open(format, bufferSize);
}
catch (LineUnavailableException e)
{
e.printStackTrace();
return;
}
line.start();
try
{
int numBytesRead = 0;
while (numBytesRead != -1)
{
numBytesRead = source.read(buffer, 0, buffer.length);
if (numBytesRead != -1)
line.write(buffer, 0, numBytesRead);
}
}
catch (IOException e){e.printStackTrace();}
line.drain();
line.close();
}
}
LOGCAT:
逻辑猫：
javax.sound.sampled.UnsupportedAudioFileException: could not get audio input stream from input file
at javax.sound.sampled.AudioSystem.getAudioInputStream(Unknown Source)
at SoundPlayer.<init>(SoundPlayer.java:19)
at Main.main(Main.java:8)
Exception in thread "main" java.lang.NullPointerException
at java.io.ByteArrayInputStream.<init>(Unknown Source)
at Main.main(Main.java:9)
In advance, thanks a lot !
提前，非常感谢！
采纳答案by Tofuw
I've found a solution to my problem. In my case, the use of JAVAZOOMlibrairy is good.
我找到了解决我的问题的方法。就我而言，JAVAZOOM库的使用很好。
Here is a sample, which only play an audio file when launching (no graphical part)
这是一个示例，它在启动时仅播放音频文件（无图形部分）
public class Sound
{
private boolean isPlaying = false;
private AdvancedPlayer player = null;
public Sound(String path) throws Exception
{
InputStream in = (InputStream)new BufferedInputStream(new FileInputStream(new File(path)));
player = new AdvancedPlayer(in);
}
public Sound(String path,PlaybackListener listener) throws Exception
{
InputStream in = (InputStream)new BufferedInputStream(new FileInputStream(new File(path)));
player = new AdvancedPlayer(in);
player.setPlayBackListener(listener);
}
public void play() throws Exception
{
if (player != null)
{
isPlaying = true;
player.play();
}
}
public void play(int begin,int end) throws Exception
{
if (player != null)
{
isPlaying = true;
player.play(begin,end);
}
}
public void stop() throws Exception
{
if (player != null)
{
player.stop();
isPlaying = false;
}
}
public boolean isPlaying()
{
return isPlaying;
}
public static void main(String[] args)
{
System.out.println("lecture de son");
try
{
Sound sound = new Sound("C:/Documents and Settings/cngo/Bureau/Stage-Save/TCPIP_AndroidJava/TCPIP_V6_Sound/OpeningSuite.mp3");
System.out.println("playing : " + sound.isPlaying());
sound.play();
System.out.println("playing : " + sound.isPlaying());
}
catch (Exception e){e.printStackTrace();}
}
}
Thanks to @murtaza.webdev for his answers !
感谢@murtaza.webdev 的回答！
回答by murtaza.webdev
Write down the full path of your music file it will works
写下您的音乐文件的完整路径，它将起作用
回答by murtaza.webdev
That exception will stay. *.wma files are not supported by standard.
该例外将保留。*.wma 文件不受标准支持。
Simplest solution would be to use *.wav files or other supported files
最简单的解决方案是使用 *.wav 文件或其他支持的文件
You can get more info on:
您可以获得更多信息：
回答by Andrew Thompson
SoundPlayer player = new SoundPlayer("C:/Documents and Settings/All Users/" +
"Documents/Ma musique/échantillons de musique/" +
"Symphonie n° 9 de Beethoven (scherzo).wma")
Ah, WMA. Great format, Java (Standard Edition) does not provide a Service Provider Interface that supports it.
啊，WMA。很好的格式，Java（标准版）不提供支持它的服务提供者接口。
You will either need to supply an SPI to allow Java Sound to support it, or use a different API. I don't know of any APIs that provide support for WMA. Can you encode it in a different format?
您将需要提供一个 SPI 以允许 Java Sound 支持它，或者使用不同的 API。我不知道提供 WMA 支持的任何 API。你能用不同的格式编码吗？
See the Java Sound info. pagefor a way to support MP3, but it requires the MP3 SPI from JMF.
请参阅Java 声音信息。有关支持 MP3 的方法的页面，但它需要来自 JMF 的 MP3 SPI。
