java - 按下按键后绘制的对象不会清除先前的矩形

Question

我的代码不会清除之前的矩形。我认为通过使用 getRect() ，然后在 keyPressed() 方法中使用 repaint() ，就可以解决问题。

import java.awt.Graphics;
import java.awt.Graphics2D;
import java.awt.event.KeyEvent;
import java.awt.geom.Rectangle2D;
import javax.swing.Icon;
import java.awt.*;
import java.awt.event.KeyEvent;
import java.awt.event.KeyListener;
import java.awt.geom.AffineTransform;
import java.awt.geom.Arc2D;
import java.awt.geom.PathIterator;
import java.awt.geom.Point2D;
import java.awt.geom.Rectangle2D;
import java.util.ArrayList;

import javax.swing.Icon;
import javax.swing.JComponent;
import javax.swing.JFrame;
import javax.swing.JLabel;
import javax.swing.JOptionPane;


public class Rectangle extends JFrame implements KeyListener{

private int height;
private int width;
private int x;
private int y;
private Graphics2D g2;



public Rectangle(int width, int height,int x, int y)
{
    this.height = height;
    this.width = width;
    this.x = x;
    this.y = y;

   JLabel label = new JLabel();


    setSize(1000,1000);
    setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
    setLocationRelativeTo(null);
    setVisible(true);


    label.addKeyListener(this);
    label.setFocusable(true);
    label.setOpaque(true);
    label.addKeyListener(this);
    this.add(label);
    setVisible(true);
    label.requestFocusInWindow();

}


public void keyPressed(KeyEvent e)
{   

    if (e.getKeyCode() == KeyEvent.VK_DOWN)
    {
        g2.clearRect(x, y, width, height);
        repaint();
        y = y + 5;
        repaint();
    }
    if (e.getKeyCode() == KeyEvent.VK_LEFT)
    {
        g2.clearRect(x, y, width, height);
        repaint();
        x = x - 5;
        repaint();
    }
    if (e.getKeyCode() == KeyEvent.VK_UP)
    {
        g2.clearRect(x, y, width, height);
        repaint();
        y = y - 5;
        repaint();
    }
    if (e.getKeyCode() == KeyEvent.VK_RIGHT)
    {
        g2.clearRect(x, y, width, height);
        x = x + 5;
        repaint();
    }
}


public void keyReleased(KeyEvent e) {}

public void keyTyped(KeyEvent e) {}




public void paint(Graphics g) {

    g2 = (Graphics2D) g;
    g2.drawRect(x, y, width, height);

}

}

测试方法:

public class Test {

public static void main(String[] args)
{
    Rectangle rectangle = new Rectangle(400,200,500,500);

}
}

Answer 1

1. setVisible(true) 应该是 public Rectangle(...) 中的最后一个代码行，因为您将 JComponent 添加到已经可见的 Swing GUI。

2. 默认情况下，JFrame 对于 KeyEvent 无法获得焦点。

3. 将JPanel放到JFrame的CENTER区域，默认:myFrame.add(myPanel).

4. 覆盖 JPanel 的 getPreferredSize，而不是设置 JFrame 大小；应通过调用 myFrame.pack() 来初始化大小。

5. JPanel 必须可聚焦才能接收 KeyEvent、setFocusable(true)；使用 KeyBindings 而不是在 JComponent 层次结构中寻找焦点。

6. 覆盖JPanel的paintComponent；第一个代码行应该是 super.paintComponent，否则会出现绘画伪像。