GUI 在 JPanel 上从动画绘制 Wave 时冻结,尽管我使用了 Swing Timer

问题描述 投票:0回答:2

请看我的代码片段,它有什么问题,当 Swing 计时器统计数据重复绘制在 jpnael 上时,它会冻结 GUI ??

class WaveformPanel extends JPanel {

        Timer graphTimer = null;
        AudioInfo helper = null;

        WaveformPanel() {
            setPreferredSize(new Dimension(200, 80));
            setBorder(BorderFactory.createLineBorder(Color.BLACK));
            graphTimer = new Timer(15, new TimerDrawing());
        }

        /**
         * 
         */
        private static final long serialVersionUID = 969991141812736791L;
        protected final Color BACKGROUND_COLOR = Color.white;
        protected final Color REFERENCE_LINE_COLOR = Color.black;
        protected final Color WAVEFORM_COLOR = Color.red;

        protected void paintComponent(Graphics g) {

            super.paintComponent(g);

            int lineHeight = getHeight() / 2;
            g.setColor(REFERENCE_LINE_COLOR);
            g.drawLine(0, lineHeight, (int) getWidth(), lineHeight);

            if (helper == null) {
                return;
            }

            drawWaveform(g, helper.getAudio(0));

        }

        protected void drawWaveform(Graphics g, int[] samples) {

            if (samples == null) {
                return;
            }

            int oldX = 0;
            int oldY = (int) (getHeight() / 2);
            int xIndex = 0;

            int increment = helper.getIncrement(helper
                    .getXScaleFactor(getWidth()));
            g.setColor(WAVEFORM_COLOR);

            int t = 0;

            for (t = 0; t < increment; t += increment) {
                g.drawLine(oldX, oldY, xIndex, oldY);
                xIndex++;
                oldX = xIndex;
            }

            for (; t < samples.length; t += increment) {
                double scaleFactor = helper.getYScaleFactor(getHeight());
                double scaledSample = samples[t] * scaleFactor;
                int y = (int) ((getHeight() / 2) - (scaledSample));
                g.drawLine(oldX, oldY, xIndex, y);

                xIndex++;
                oldX = xIndex;
                oldY = y;
            }
        }

        public void setAnimation(boolean turnon) {
            if (turnon) {
                graphTimer.start();
            } else {
                graphTimer.stop();
            }
        }

        class TimerDrawing implements ActionListener {

            @Override
            public void actionPerformed(ActionEvent e) {

                byte[] bytes = captureThread.getTempBuffer();

                if (helper != null) {
                    helper.setBytes(bytes);
                } else {
                    helper = new AudioInfo(bytes);
                }
                repaint();
            }
        }

    }

我正在从其父类调用 WaveFormPanel 的 setAnimation。当动画开始时,它不会绘制任何东西,但会冻结。请给我解决方案。

谢谢 米希尔·帕雷克

java swing timer bufferedimage paintcomponent
2个回答
4
投票

java.swingx.Timer
ActionPerformed
中调用
EDT
。接下来的问题是,什么需要时间来渲染。这可能是对
captureThread.getTempBuffer
的调用,也可能是帮助的构建,但我怀疑这只是您要绘制的大量数据。

最近在玩这个,处理波形需要相当多的时间

一个建议可能是减少您绘制的样本数量。与其绘制每个样本点,不如根据组件的宽度每隔一秒或第四个样本点绘制一次。你仍然应该得到同样的东西,但没有所有的工作......

已更新

所有样本,2.18 秒

AllSamples

每 4 个样本,0.711 秒

enter image description here

每 8 个样本,0.450 秒

enter image description here

然后根据计时器进行绘制,也许您需要根据批量数据进行绘制。

由于您的加载程序线程有数据“块”,因此可能会绘制它。

正如 HoverCraftFullOfEels 所建议的那样,您可以先将其绘制到 BufferedImage,然后再将其绘制到屏幕上...

SwingWorker 或许可以为您实现这个

已更新

这是我用来绘制上述样本的代码。

// Samples is a 2D int array (int[][]), where the first index is the channel, the second is the sample for that channel
if (samples != null) {

    Graphics2D g2d = (Graphics2D) g;

    int length = samples[0].length;

    int width = getWidth() - 1;
    int height = getHeight() - 1;

    int oldX = 0;
    int oldY = height / 2;
    int frame = 0;

    // min, max is the min/max range of the samples, ie the highest and lowest samples
    int range = max + (min * -2);
    float scale = (float) height / (float) range;

    int minY = Math.round(((height / 2) + (min * scale)));
    int maxY = Math.round(((height / 2) + (max * scale)));

    LinearGradientPaint lgp = new LinearGradientPaint(
            new Point2D.Float(0, minY),
            new Point2D.Float(0, maxY),
            new float[]{0f, 0.5f, 1f},
            new Color[]{Color.BLUE, Color.RED, Color.BLUE});
    g2d.setPaint(lgp);
    for (int sample : samples[0]) {

        if (sample % 64 == 0) {

            int x = Math.round(((float) frame / (float) length) * width);
            int y = Math.round((height / 2) + (sample * scale));

            g2d.drawLine(oldX, oldY, x, y);

            oldX = x;
            oldY = y;

        }

        frame++;

    }

}

我使用

AudioStream
流加载 Wav 文件并生成 2D 样本。


3
投票

我猜您的波形图代码(从

paintComponent(...)
方法中调用)花费的时间比您想象的要长,并且占用了 Swing 绘画和 EDT。

如果这是我的代码,我会考虑将我的波浪绘制到 BufferedImages once,从这些图像制作 ImageIcons,然后简单地交换我的 Swing Timer 中的图标。

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