Java Swing - 将矩形拖到 JPanel 上的有效方法?

标签 java swing user-interface drawing

我有一些绘制矩形的代码。它用于在 JPanel 上绘制矩形,以标记小部件的边界。首先是代码,然后我将解释我的问题 cq。问题。

首先,我有一个扩展 JPanel 的类 (WidgetDrawingPanel)。

public WidgetDrawingPanel(int width, int height) {
    /*To make things visible at least*/
    widgets.add(new Widget(10,10,100,100, WidgetType.TextField));
    widgets.add(new Widget(50,50,100,200, WidgetType.TextField));
    this.width = width;
    this.height = height;
    this.setBackground(Color.BLUE);
    addListener(); //adds both MouseMotionListener and MouseListener
}

下面你会看到我经常引用 ch。这是一个 CoordinateHolder,它保存我的鼠标移动的开始坐标和当前坐标。

private void addListener() {
    this.addMouseMotionListener(new MouseMotionListener() {
        @Override
        public void mouseDragged(MouseEvent arg0) {
            ch.currentX = arg0.getX();
            ch.currentY = arg0.getY();
            System.out.println("dragging " + ch.currentX + ","+ch.currentY);
            WidgetDrawingPanel.this.repaint();
        }
    });
    this.addMouseListener(new MouseListener() {
        @Override
        public void mouseReleased(MouseEvent event) {
            ch.endX = event.getX();
            ch.endY = event.getY();
            try {
                checkCoords();
            } catch (OutsidePanelException e) {
                e.printStackTrace();
                JOptionPane.showMessageDialog(null, "drawn Outside Panel");
            }
        }

        @Override
        public void mousePressed(MouseEvent event) {
            ch = new CoordinateHolder(event.getX(), event.getY());
        }
    });
}

最后是 paintComponent(Grapics) 方法。里面有遍历Widgets,里面其实只是已经绘制好的Rects(x,y,w,h属性),只是里面的信息多了一点,这里面用处不大应用程序的绘图部分。每次释放鼠标时,CoordinateHolder 都会转换为 Widget,并添加到 widgets

@Override
public void paintComponent(Graphics g) {
    super.paintComponent(g);
    System.out.println("Paint");
    g.setColor(Color.BLUE);
    g.fillRect(0, 0, width, height); //making the whole panel blue
    g.setColor(Color.RED);
    Graphics2D g2 = (Graphics2D)g;
    g2.setStroke(new BasicStroke(3));
    for (Widget w : widgets) { 
        g.drawRect(w.getX(), w.getY(), w.getW(), w.getH());
    }
    if (ch != null)
        g.drawRect(ch.startX, ch.startY, ch.currentX - ch.startX, ch.currentY - ch.startY);
}

这段代码有效,但我怀疑这是非常低效和低效的,因为上面的代码在鼠标拖动时不断刷新 JPanel,也就是说,每 10 毫秒刷新一次?我想它很快就会变慢,尤其是当用户创建大量矩形时(它们也会不断重绘,如 painComponent(Graphics) 所示)。

问题 cq。问题

是否有更好、资源消耗更少的方法,用户可以在其中平滑地拖动矩形?

我阅读了这个 Drag rectangle on JFrame in Java 的答案,但该答案的作者似乎和我一样。但同样,那是表现不佳的方式,对吗?或者计算机是否应该能够轻松地不断重绘组件,这实际上是一种有效的方法吗?

最佳答案

要显示大量不变的背景形状,请将它们绘制到 BufferedImage,然后在 paintComponent(...) 方法中显示该 BufferedImage。因此,在绘制形状时,在 paintComponent(...) 中绘制它,但是一旦绘制完成,可能是在 mouseRelease 上,然后在背景 BufferedImage 中绘制它。

请注意,最能减慢您当前绘图代码的可能是您的调试 SOP 语句,但我认为这些将从完成的代码中删除。

例如:

import java.awt.*;
import java.awt.event.*;
import java.awt.image.BufferedImage;

import javax.swing.*;

@SuppressWarnings("serial")
public class DrawingPanel extends JPanel {
   private static final int PREF_W = 600;
   private static final int PREF_H = 400;
   private static final Color DRAWING_COLOR = new Color(255, 100, 200);
   private static final Color FINAL_DRAWING_COLOR = Color.red;

   private BufferedImage backgroundImg;
   private Point startPt = null;
   private Point endPt = null;
   private Point currentPt = null;

   public DrawingPanel() {
      backgroundImg = new BufferedImage(PREF_W, PREF_H,
            BufferedImage.TYPE_INT_ARGB);
      Graphics g = backgroundImg.getGraphics();
      g.setColor(Color.blue);
      g.fillRect(0, 0, PREF_W, PREF_H);
      g.dispose();

      MyMouseAdapter myMouseAdapter = new MyMouseAdapter();
      addMouseMotionListener(myMouseAdapter);
      addMouseListener(myMouseAdapter);
   }

   @Override
   protected void paintComponent(Graphics g) {
      super.paintComponent(g);
      if (backgroundImg != null) {
         g.drawImage(backgroundImg, 0, 0, this);
      }

      if (startPt != null && currentPt != null) {
         g.setColor(DRAWING_COLOR);
         int x = Math.min(startPt.x, currentPt.x);
         int y = Math.min(startPt.y, currentPt.y);
         int width = Math.abs(startPt.x - currentPt.x);
         int height = Math.abs(startPt.y - currentPt.y);
         g.drawRect(x, y, width, height);
      }
   }

   @Override
   public Dimension getPreferredSize() {
      return new Dimension(PREF_W, PREF_H);
   }

   public void drawToBackground() {
      Graphics g = backgroundImg.getGraphics();
      g.setColor(FINAL_DRAWING_COLOR);
      int x = Math.min(startPt.x, endPt.x);
      int y = Math.min(startPt.y, endPt.y);
      int width = Math.abs(startPt.x - endPt.x);
      int height = Math.abs(startPt.y - endPt.y);
      g.drawRect(x, y, width, height);
      g.dispose();

      startPt = null;
      repaint();
   }

   private class MyMouseAdapter extends MouseAdapter {
      @Override
      public void mouseDragged(MouseEvent mEvt) {
         currentPt = mEvt.getPoint();
         DrawingPanel.this.repaint();
      }

      @Override
      public void mouseReleased(MouseEvent mEvt) {
         endPt = mEvt.getPoint();
         currentPt = null;
         drawToBackground();
      }

      @Override
      public void mousePressed(MouseEvent mEvt) {
         startPt = mEvt.getPoint();
      }
   }

   private static void createAndShowGui() {
      DrawingPanel mainPanel = new DrawingPanel();

      JFrame frame = new JFrame("Drawing Panel");
      frame.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
      frame.getContentPane().add(mainPanel);
      frame.pack();
      frame.setLocationByPlatform(true);
      frame.setVisible(true);
   }

   public static void main(String[] args) {
      SwingUtilities.invokeLater(new Runnable() {
         public void run() {
            createAndShowGui();
         }
      });
   }
}

关于Java Swing - 将矩形拖到 JPanel 上的有效方法?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/15312070/

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