当我在JPanel中运行模拟移动方块时,GUI不会更新。在整个模拟完成之前,它不会刷新JPanel。
我的代码:
import javax.swing.SwingUtilities;
import javax.swing.JFrame;
import javax.swing.JPanel;
import javax.swing.BorderFactory;
import java.awt.Color;
import java.awt.Dimension;
import java.awt.Graphics;
public class Main {
public static void main(String[] args) {
SwingUtilities.invokeLater(new Runnable() {
public void run() {
createAndShowGUI();
}
});
}
private static void createAndShowGUI() {
System.out.println("Created GUI on EDT? "+
SwingUtilities.isEventDispatchThread());
JFrame f = new JFrame("Swing Paint Demo");
f.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
f.add(new MyPanel());
f.pack();
f.setVisible(true);
}
}
class MyPanel extends JPanel {
private static final long serialVersionUID = 1L;
private int squareX = 50;
private int squareY = 50;
private int squareW = 20;
private int squareH = 20;
public MyPanel() {
setBorder(BorderFactory.createLineBorder(Color.black));
doAnimation();
}
private void doAnimation(){
SwingUtilities.invokeLater(new Runnable() {
public void run() {
for(int x = 0; x < 250-squareW; x+=squareX){
for(int y = 0; y < 200-squareH; y+=squareY){
moveSquare(x,y);
}
}
}
});
}
private void moveSquare(int x, int y) {
int OFFSET = 1;
if ((squareX!=x) || (squareY!=y)) {
repaint(squareX,squareY,squareW+OFFSET,squareH+OFFSET);
squareX=x;
squareY=y;
repaint(squareX,squareY,squareW+OFFSET,squareH+OFFSET);
}
}
public Dimension getPreferredSize() {
return new Dimension(250,200);
}
protected void paintComponent(Graphics g) {
super.paintComponent(g);
g.drawString("This is my custom Panel!",10,20);
g.setColor(Color.RED);
g.fillRect(squareX,squareY,squareW,squareH);
g.setColor(Color.BLACK);
g.drawRect(squareX,squareY,squareW,squareH);
}
}
答案 0 :(得分:4)
在完成整个模拟之前,它不会刷新JPanel 完成。
是。更新将执行一次。因为您在invokeLater
的Runnable中运行以下代码,该代码通过包装为单个事件提交给EDT:
SwingUtilities.invokeLater(new Runnable() {
public void run() {
for(int x = 0; x < 250-squareW; x+=squareX){
for(int y = 0; y < 200-squareH; y+=squareY){
moveSquare(x,y);
}
}
}
});
您在moveSquare(x, y)
方法中创建的所有重绘请求都包含在该单个事件中。即使我们在同一个事件处理程序中连续几次调用重绘,Swing也足够聪明,可以在一次绘制操作中获取该信息并重新绘制屏幕的这些部分。换句话说,Swing不会连续多次重新绘制该组件,即使这是代码似乎正在执行的操作。请查看this official tutorial page以获取详细说明和示例。
然而,运行这样的长循环将使您的GUI应用程序无响应:因为它可能等待EDT。并利用Swing Timer
类和周期性任务来制作动画效果。
答案 1 :(得分:-1)
看这里..
private void moveSquare(int x, int y) {
int OFFSET = 1;
if ((squareX!=x) || (squareY!=y)) {
repaint(squareX,squareY,squareW+OFFSET,squareH+OFFSET);
squareX=x; // It makes squareX will be 0 in first iteration
squareY=y; // It makes squareY will be 0 in first iteration
repaint(squareX,squareY,squareW+OFFSET,squareH+OFFSET);
}
}
最后你得到了无限循环。所以改变它..