无法同时播放声音

Question

我试图通过单击j按钮同时播放六个音频轨道，但是单击时会播放第一个轨道，并等待直到完成播放第二个轨道，依此类推。 这是我的代码

 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);
            }
        }
    });

和音频播放器导入

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);
}}

在播放曲目时，该按钮也保持单击状态，因此我也无法使用音量jslider。 谢谢您的帮助！

Answer 1

您编写play方法的方式将阻塞直到流完全播放为止-这意味着流将一个接一个地播放。 一种选择是为每个流派生一个新线程。 这样可以避免阻塞问题，但是会带来另一个问题，那就是线程都将处于启动状态。 这意味着这些流不一定都必须在完全相同的时间开始（尽管您可以使用信号使它们非常接近同步）。

我认为一种更好的方法是使用从所有文件读取并在单个线程中全部写入一个 SourceDataLine的方法。 这意味着您必须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，这并不是很难。 我假设使用16位样本。 如果您的文件不同，则可以弄清楚如何处理。

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;
        }
    }
}

并通过传递文件名数组来调用它（我建议将其替换为track1，track2等...）

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

第三种可能更好的选择是从InputStream派生一个支持多个文件并在内部进行混合的类。 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。 这比我现在想参与的要复杂得多。

PS我没有尝试编译任何。 我只是想传达这个想法。