用Java一个接一个地播放WAV文件

时间:2010-06-22 13:46:04

标签: java audio concatenation wav sequential

我正在尝试互相播放一些WAV个文件。我试过这个方法:

for (String file : audioFiles) {
    new AePlayWave(file).start();
}

但是,它同时播放所有这些。所以我需要一个看起来像这样的函数:

public void play(Vector<String> audioFiles);

向量包含文件,例如:"test1.wav""test2.wav"

我一直在寻找超过四个小时,但我似乎无法找到一个有效的解决方案:(

我还尝试将WAV文件连接到一个AudioInputStream。它没有给出任何编译器错误,但声音完全搞砸了。代码:

public static AudioInputStream concat(Vector<String> files) throws UnsupportedAudioFileException, IOException {
    AudioInputStream total = AudioSystem.getAudioInputStream(new File(files.get(0)));

    for (int i = 1; i < files.size(); i++) {
        AudioInputStream clip = AudioSystem.getAudioInputStream(new File(files.get(i)));
        total = new AudioInputStream(new SequenceInputStream(total, clip),
                                     total.getFormat(),
                                     total.getFrameLength() + clip.getFrameLength());
    }
    return total;
}

修改

即使我尝试将两个第一个文件放在一起,也会失败:

public static AudioInputStream concat(Vector<String> files) throws UnsupportedAudioFileException, IOException {
    AudioInputStream clip1 = AudioSystem.getAudioInputStream(new File(files.get(0)));
    AudioInputStream clip2 = AudioSystem.getAudioInputStream(new File(files.get(1)));

    AudioInputStream total = new AudioInputStream(
        new SequenceInputStream(clip1, clip2),
        clip1.getFormat(),
        clip1.getFrameLength() + clip2.getFrameLength());

    return total;
}

2 个答案:

答案 0 :(得分:2)

这段代码有点低级,但有效:

    byte[] buffer = new byte[4096];
    for (File file : files) {
        try {
            AudioInputStream is = AudioSystem.getAudioInputStream(file);
            AudioFormat format = is.getFormat();
            SourceDataLine line = AudioSystem.getSourceDataLine(format);
            line.open(format);
            line.start();
            while (is.available() > 0) {
                int len = is.read(buffer);
                line.write(buffer, 0, len);
            }
            line.drain(); //**[DEIT]** wait for the buffer to empty before closing the line
            line.close();
        } catch (Exception e) {
            e.printStackTrace();
        }
    }

基本上你打开一个AudioInputStream,读取数据并将其写入SourceDataLine。 write方法阻塞,因此它将播放文件。

您可以尝试将Clip用于相同目的。

答案 1 :(得分:0)

最终答案(排水):

public static void play(ArrayList<String> files){
    byte[] buffer = new byte[4096];
    for (String filePath : files) {
        File file = new File(filePath);
        try {
            AudioInputStream is = AudioSystem.getAudioInputStream(file);
            AudioFormat format = is.getFormat();
            SourceDataLine line = AudioSystem.getSourceDataLine(format);
            line.open(format);
            line.start();
            while (is.available() > 0) {
                int len = is.read(buffer);
                line.write(buffer, 0, len);
            }
            line.drain();
            line.close();
        } catch (Exception ex) {
            ex.printStackTrace();
        }
    }
}