Java 无法同时播放声音

我试图在点击一个按钮的同时播放六个音轨，但点击后它会播放第一个音轨并等待播放完第二个音轨，依此类推。这是我的密码

 button.addActionListener(new ActionListener() {
        public void actionPerformed(ActionEvent e) {
            if (e.getSource() == button) {
                System.out.println("Button Pressed");
                AudioPlayerExample2 player1 = new AudioPlayerExample2();
                AudioPlayerExample2 player2 = new AudioPlayerExample2();
                AudioPlayerExample2 player3 = new AudioPlayerExample2();
                AudioPlayerExample2 player4 = new AudioPlayerExample2();
                AudioPlayerExample2 player5 = new AudioPlayerExample2();
                AudioPlayerExample2 player6 = new AudioPlayerExample2();
                player1.play(track1);
                player2.play(track2);
                player3.play(track3);
                player4.play(track4);
                player5.play(track5);
                player6.play(track6);
            }
        }
    });

还有音频播放器

public class AudioPlayerExample2 {

private static final int BUFFER_SIZE = 4096;


public void play(String audioFilePath) {
    File audioFile = new File(audioFilePath);
    try {
        AudioInputStream audioStream = AudioSystem.getAudioInputStream(audioFile);

        AudioFormat format = audioStream.getFormat();

        DataLine.Info info = new DataLine.Info(SourceDataLine.class, format);

        SourceDataLine audioLine = (SourceDataLine) AudioSystem.getLine(info);

        audioLine.open(format);

        audioLine.start();

        System.out.println("Playback started.");

        byte[] bytesBuffer = new byte[BUFFER_SIZE];
        int bytesRead = -1;

        while ((bytesRead = audioStream.read(bytesBuffer)) != -1) {
            audioLine.write(bytesBuffer, 0, bytesRead);
        }

        audioLine.drain();
        audioLine.close();
        audioStream.close();

        System.out.println("Playback completed.");

    } catch (UnsupportedAudioFileException ex) {
        System.out.println("The specified audio file is not supported.");
        ex.printStackTrace();
    } catch (LineUnavailableException ex) {
        System.out.println("Audio line for playing back is unavailable.");
        ex.printStackTrace();
    } catch (IOException ex) {
        System.out.println("Error playing the audio file.");
        ex.printStackTrace();
    }
}

public static void main(String[] args) {
    String audioFilePath = "";
    AudioPlayerExample2 player = new AudioPlayerExample2();
    player.play(audioFilePath);
}}

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播放曲目时，按钮也会保持单击状态，因此我也无法使用我的volume jslider。谢谢你的帮助

您编写播放方法的方式将阻止播放，直到流完全播放完毕，这意味着流将一个接一个地播放。一种选择是为每个流分叉一个新线程。这将避免阻塞问题，但会引入另一个问题，即所有线程都将竞相启动。这意味着流不一定都在同一时间启动，尽管您可以使用一个信号使它们非常接近同步

我认为更好的方法是在一个线程中使用从所有文件读取和写入一个SourceDataLine。这意味着您必须自己手动将信号混合在一起。假设所有文件都具有相同的采样率和位深度，这并不太困难。我假设了16位样本。如果您的文件不同，那么您可以找到处理方法

public void play(String[] audioFilePath) {
    int numStreams = audioFilePath.length;

    // Open all of the file streams
    AudioInputStream[] audioStream = new AudioInputStream[numStreams];
    for (int i = 0; i < numStreams; i++)
    audioStream[i] = AudioSystem.getAudioInputStream(audioFile);

    // Open the audio line.
    AudioFormat format = audioStream[0].getFormat();
    DataLine.Info info = new DataLine.Info(SourceDataLine.class, format);
    SourceDataLine audioLine = (SourceDataLine) AudioSystem.getLine(info);
    audioLine.open(format);
    audioLine.start();

    while (true) {
        // Read a buffer from each stream and mix into an array of
        // doubles.
        byte[] bytesBuffer = new byte[BUFFER_SIZE];
        double[] mixBuffer = new double[BUFFER_SIZE/2];
        int maxSamplesRead = -1;
        for (int i = 0 ; i < numStreams; i++)
        {
            int bytesRead = audioStream.read(bytesBuffer);
            if (bytesRead != -1) {
                int samplesRead = bytesRead/2;
                if (samplesRead > maxSamplesRead) {
                    maxSamplesRead = samplesRead;
                }
                for (int j = 0 ; j < bytesRead/2 ; j++) {
                    double sample = ((bytesBuffer[j*2] << 8) | bytesBuffer[j*2+1]) / 32768.0;
                    mixBuffer[j] += sample;
                }
            }
        }

        // Convert the mixed samples back into a byte array and play.
        if (maxSamplesRead > 0) {
            for (int i = 0; i < maxSamplesRead; i++) {
                // rescale data between -1 and 1
                mixBuffer[i] /= numStreams;

                // and now back to 16-bit
                short sample16 = (short)(mixBuffer * 32768);

                // and back to bytes
                bytesBuffer[i*2]   = (byte)(sample16 >> 8);
                bytesBuffer[i*2+1] = (byte)(sample16);
            }
            audioLine.write(bytesBuffer, 0, maxSamplesRead*2);
        }
        else {
            // All of the streams are empty so cleanup.
            audioLine.drain();
            audioLine.close();
            for (int i = 0 ; i < numStreams; i++)
                audioStream[i].close();
            break;
        }
    }
}

第三种可能更好的选择是从InputStream派生一个类，该类支持多个文件并在内部进行混合。通过这种方法，您可以使用大多数现有的AudioPlayerExample2类，但只有一个实例。这件事会比我现在想做的要复杂得多

---

顺便说一句，我还没试过编译这些东西。我只是想让大家明白这个想法。

排水管堵塞不是吗？我觉得你似乎很难做到这一点，你可以简单地使用剪辑和它的开始，停止和其他功能，正如播放曲目时所演示的那样，按钮也会保持单击状态，因此我也无法使用我的音量jslider。-这意味着您正在阻止EDT并阻止更新/重新绘制UI或响应UI事件，这可能与您的第一个问题有关。你需要找到你正在使用的哪些方法阻止了当前线程，因为你的播放方法在曲目播放完成之前不会返回？我不能使用Clip，因为音频文件非常大，为50mb和@immibis，我如何解决这个问题？

button.addActionListener(new ActionListener() {
        public void actionPerformed(ActionEvent e) {
            if (e.getSource() == button) {
                System.out.println("Button Pressed");
                AudioPlayerExample2 player = new AudioPlayerExample2(allTracks);
            }
        }
    });