Question

public class Graphics2DTest extends JPanel implements ActionListener{
private Timer time = new Timer(5,(ActionListener) this);
int x = 0,y = 0;
public void paintComponent(Graphics g){

    Graphics2D gui = (Graphics2D) g;
    Rectangle2D rectangle = new Rectangle2D.Double(x,y,100,150);
    gui.setPaint(Color.GREEN);
    gui.fill(rectangle);
    time.start();
}

public void actionPerformed(ActionEvent arg0) {
    x++;
    y++;
    repaint();
}
}

问题是repaint（）应该清除框架并在该位置绘制矩形，但之前绘制的矩形仍然存在。那么，怎么做呢？请解释一下你的答案。

Answer 1

您是否尝试在paintComponent方法中调用super.paintComponent（g）？这将清除JPanel中绘制的先前图像：

public void paintComponent(Graphics g){
  super.paintComponent(g);
  Graphics2D gui = (Graphics2D) g;
  Rectangle2D rectangle = new Rectangle2D.Double(x,y,100,150);
  gui.setPaint(Color.GREEN);
  gui.fill(rectangle);
  //time.start();
}

此外，不要在paintComponent方法中启动计时器或执行任何程序逻辑。首先，你无法绝对控制何时或是否会调用该方法，其次，这种方法必须只涉及绘画而不是其他任何东西，并且需要尽可能快。

例如：

import java.awt.*;
import java.awt.event.*;
import java.awt.geom.*;

import javax.swing.*;

public class Graphics2DTest extends JPanel implements ActionListener {
    private Timer time = new Timer(5, (ActionListener) this);
    int x = 0, y = 0;

    public void paintComponent(Graphics g) {
        super.paintComponent(g);
        Graphics2D gui = (Graphics2D) g;
        Rectangle2D rectangle = new Rectangle2D.Double(x, y, 100, 150);
        gui.setPaint(Color.GREEN);
        gui.fill(rectangle);
        //time.start();
    }

    public void actionPerformed(ActionEvent arg0) {
        x++;
        y++;
        repaint();
    }

    public Graphics2DTest() {
        setPreferredSize(new Dimension(700, 500));
        time.start();
    }

    private static void createAndShowUI() {
        JFrame frame = new JFrame("Graphics2DTest");
        frame.getContentPane().add(new Graphics2DTest());
        frame.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
        frame.pack();
        frame.setLocationRelativeTo(null);
        frame.setVisible(true);
    }

    public static void main(String[] args) {
        java.awt.EventQueue.invokeLater(new Runnable() {
            public void run() {
                createAndShowUI();
            }
        });
    }
}

Answer 2

您每次都需要重新绘制背景。在绘制矩形之前添加代码以绘制背景。

Answer 3

您需要先清除背景。

资源就是这样：

http://java.sun.com/products/jfc/tsc/articles/painting/

Repaint（）不会清除帧

3 个答案: