Question

我正在尝试制作一个JButton 启动一个解决迷宫的动画。我有一个递归的，深度优先的搜索函数，它找到路径，但逐步着色（使用Thread.sleep()）。代码完美无缺。我遇到的问题是有一个按钮开始这个动画。

当我将函数添加到JButton的actionPerformed中时，解决方案只有在完全完成后才会出现（不是一步一步）。我得知这是因为动作侦听器是单线程或类似的东西，所以我尝试使用Swing Timer。我只是创建了一个每隔几毫秒重新绘制JPanel的事件，认为这会在后台重新绘制JPanel，从而显示找到解决方案的进度。这给了我一个运行时错误：

线程“AWT-EventQueue-0”中的异常java.lang.NullPointerException 在游戏$ 1.actionPerformed（Game.java:54）在javax.swing.Timer.fireActionPerformed（Timer.java:313）在javax.swing.Timer $ DoPostEvent.run（Timer.java:245） at java.awt.event.InvocationEvent.dispatch（InvocationEvent.java:311） at java.awt.EventQueue.dispatchEventImpl（EventQueue.java:756） at java.awt.EventQueue.access $ 500（EventQueue.java:97） at java.awt.EventQueue $ 3.run（EventQueue.java:709） at java.awt.EventQueue $ 3.run（EventQueue.java:703） at java.security.AccessController.doPrivileged（Native Method） at java.security.ProtectionDomain $ JavaSecurityAccessImpl.doIntersectionPrivilege（ProtectionDomain.java:76） at java.awt.EventQueue.dispatchEvent（EventQueue.java:726） at java.awt.EventDispatchThread.pumpOneEventForFilters（EventDispatchThread.java:201） at java.awt.EventDispatchThread.pumpEventsForFilter（EventDispatchThread.java:116） at java.awt.EventDispatchThread.pumpEventsForHierarchy（EventDispatchThread.java:105） at java.awt.EventDispatchThread.pumpEvents（EventDispatchThread.java:101） at java.awt.EventDispatchThread.pumpEvents（EventDispatchThread.java:93）在java.awt.EventDispatchThread.run（EventDispatchThread.java:82）

关于如何从JButton启动动画的任何想法？我觉得这应该比我制作它简单得多，因为动画本身就可以正常运行，而我想要的只是点击一个JButton。我以为我会从零代码开始，并根据要求提供所需的一切。

Game.java

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

public class Game {
    private static MazeGrid scene;
    private static JFrame frame;
    private static JButton play;
    private static JButton solve;
    private static JButton exit;
    private static Timer timer;
    private static boolean s = false;


    public static void init() {
        scene = new MazeGrid();
        frame = new JFrame();
        play = new JButton("Play");
        solve = new JButton("Solve");
        exit = new JButton("Exit");
        frame.setResizable(false);
        frame.setSize(900, 700);
        frame.setLayout(null);
        frame.setLocationRelativeTo(null);

        scene.setSize(650, 680);
        frame.add(scene, BorderLayout.CENTER);
        play.setSize(100, 50);
        play.setLocation(650, 10);
        frame.add(play);
        solve.setSize(100, 50);
        solve.setLocation(770, 10);
        frame.add(solve);
        exit.setSize(100, 50);
        exit.setLocation(650, 100);
        frame.add(exit);
        frame.setVisible(true);
    }


    public static void main(String[] args) throws InterruptedException {
        ActionListener action = new ActionListener(){
            @Override
            public void actionPerformed(ActionEvent ev) {
                if (ev.getSource() == timer) {
                    scene.repaint();
                    System.out.println("hello");
                }

            }
        };

        timer = new Timer(40, action);
        timer.setInitialDelay(0);
        timer.start();

        init();
        play.addActionListener(new ActionListener() {
            @Override
            public void actionPerformed(ActionEvent e) {
                frame.setVisible(false);
                init();
            }
        });

        solve.addActionListener(new ActionListener() {
            @Override
            public void actionPerformed(ActionEvent e) {
                scene.solve(1, 1, squares, 40);
                timer.stop();
            }
        });

        exit.addActionListener(new ActionListener() {
            @Override
            public void actionPerformed(ActionEvent e) {
                frame.setVisible(false);
                frame.dispose();
            }
        });
    }

}

解决

private void solve(int x, int y, Squares squares, int delay) {
    // reached middle
    if (x == N / 2 && y == N / 2)
        done = true;
    if (x == 0 || y == 0 || x == N + 1 || y == N + 1)
        return;
    if (done || m.Visited(x, y))
        return;

    m.setVisited(x, y, true);

    if (!(x == 1 & y == 1)) {
        squares.getSquare(x, y).setPath(new Color(180, 140, 50));
        try {
            Thread.sleep(delay);
        } catch (InterruptedException e) {
            System.err.println("Error while sleeping!");
            Thread.currentThread().interrupt();
        }

        repaint();
    }

    if (!m.Top(x, y))
        solve(x, y + 1, squares, delay);
    if (!m.Right(x, y))
        solve(x + 1, y, squares, delay);
    if (!m.Bottom(x, y))
        solve(x, y - 1, squares, delay);
    if (!m.Left(x, y))
        solve(x - 1, y, squares, delay);

    if (done)
        return;
    if (!(x == 1 & y == 1)) {
        squares.getSquare(x, y).setPath(new Color(230, 230, 230));
        try {
            Thread.sleep(delay);
        } catch (InterruptedException e) {
            System.err.println("Error while sleeping!");
            Thread.currentThread().interrupt();
        }
        repaint();
    }
}

Answer 1

在您启动init()之前致电Timer，或者scene对象在尝试访问时null。{/ p>

     init();

     ActionListener action = new ActionListener(){
                @Override
                public void actionPerformed(ActionEvent ev) {
                    if (ev.getSource() == timer) {
                        scene.repaint();
                        System.out.println("hello");
                    }

                }
            };

     timer = new Timer(40, action);
     timer.setInitialDelay(0);
     timer.start();

Answer 2

我使用像这样的SwingWorker达到了我想要的效果：

class Solver  extends SwingWorker<Integer, Integer> {
    int delay;

    Solver (int delay) {
        this.delay = delay;
    }

    protected Integer doInBackground() throws Exception {
        scene.solve(delay);
        return 1;
    }
}

并将其添加到JButton：

@Override
public void actionPerformed(ActionEvent e) {
    new Solver(40).execute();
}

感谢你们的帮助。

创建一个逐步更新JPanel的ActionListener

2 个答案: