在并发循环中在计时器上更新JButton

Question

在并发循环中，让JButton重复更新(与计时器一起使用)有一些问题。我正在做一个简单的游戏，在一个10 * 10的瓷砖对象网格上玩，它对应着一个带有100个按钮的JButton arrayList。

程序的这一部分处理简单的路径查找(即，如果我单击字符，然后是一个空的瓷砖，字符将在其到达目的地的过程中通过每个瓷砖)。每个步骤之间都有一个延迟，这样用户就可以看到字符的进度。

在当前的情况下，移动是正确的，但是只有当字符到达目标时才更新JButton，而不是中间的步骤。

public void move(int terrainTile)
{
    int currentPosition = actorList.get(selectedActor).getPosition();
    int movementValue = 0;
    int destination = terrainTile;
    int destinationX = destination / 10;
    int destinationY = destination % 10;

    do
    {
        currentPosition = actorList.get(selectedActor).getPosition();  // Gets PC's current position (before move)
        System.out.println("Old position is " + currentPosition);
        int currentX = currentPosition / 10;
        int currentY = currentPosition % 10;


        if(actorList.get(selectedActor).getCurrentAP() > 0)
        {
            movementValue = 0;

            if(destinationX > currentX)
            {
                movementValue += 10;
            }

            if(destinationX < currentX)
            {
                movementValue -= 10;
            }

            if(destinationY > currentY)
            {
                movementValue += 1;
            }

            if(destinationY < currentY)
            {
                movementValue -= 1;
            }

            int nextStep = currentPosition + movementValue;

            myGame.setActorIdInTile(currentPosition, -1);  //Changes ActorId in PC current tile back to -1
            scrubTiles(currentPosition);

            actorList.get(selectedActor).setPosition(nextStep);  //  Sets new position in actor object
            System.out.println("Actor " + selectedActor + " " + actorList.get(selectedActor).getName() + " position has been updated to " + nextStep);

            myGame.setActorIdInTile(nextStep, selectedActor); // Sets ActorId in moved to Tile
            System.out.println("Tile " + nextStep + " actorId has been updated to " + selectedActor);

            buttons.get(nextStep).setIcon(new ImageIcon(actorList.get(selectedActor).getImageName()));

            // If orthagonal move AP-4
            if(movementValue == 10 || movementValue == -10 || movementValue == 1 || movementValue == -1)
            {
                actorList.get(selectedActor).reduceAP(4);
            }
            // If diagonal move AP-6
            else
            {
                actorList.get(selectedActor).reduceAP(6);
            }

            System.out.println(actorList.get(selectedActor).getName() + " has " + actorList.get(selectedActor).getCurrentAP() + " AP remaining");

            try
            {
                Thread.sleep(500);    // one second
            }
            catch (Exception e){} 

            buttons.get(nextStep).repaint();

        }

        else
        {
            System.out.println(actorList.get(selectedActor).getName() + " has insufficient AP to move");
            break;
        }

    }while(destination != (currentPosition + movementValue));

我试过的是：

1. buttons.get(nextStep).repaint()；(尝试在设置imageIcon后放置一个命令来重新绘制按钮。No change.
2. buttons.get(nextStep).revalidate()；(No 100%肯定这是一种潜在的解决方案，但不起作用。
3. 步骤1&2 combined
4. Looked进入swing计时器类-不是每次触发actionEvent时都会发生移动(只有当字符被选中并且目标瓷砖为空时)，所以不确定如何才能使其工作

。

任何帮助都将不胜感激！

Answer 1

我真的不知道你在评论中到底想知道什么，尽管上面的答案是+1，但在我看来，这才是真正的原因。看看这个示例程序，只需在move(...)中添加对timerAction方法的调用，这似乎对您是有效的。在这里，尝试下面的代码：

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

public class GridExample
{
    private static final int SIZE = 36;
    private JButton[] buttons;
    private int presentPos;
    private int desiredPos;
    private Timer timer;
    private Icon infoIcon = 
                UIManager.getIcon("OptionPane.informationIcon");

    private ActionListener timerAction = new ActionListener()
    {
        public void actionPerformed(ActionEvent ae)
        {
            buttons[presentPos].setIcon(null);
            if (desiredPos < presentPos)
            {
                presentPos--;
                buttons[presentPos].setIcon(infoIcon);
            }
            else if (desiredPos > presentPos)
            {
                presentPos++;
                buttons[presentPos].setIcon(infoIcon);
            }
            else if (desiredPos == presentPos)
            {
                timer.stop();
                buttons[presentPos].setIcon(infoIcon);
            }
        }
    };

    public GridExample()
    {
        buttons = new JButton[SIZE];
        presentPos = 0;
        desiredPos = 0;
    }

    private void createAndDisplayGUI()
    {
        JFrame frame = new JFrame("Grid Game");
        frame.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);

        JPanel contentPane = new JPanel();
        contentPane.setLayout(new GridLayout(6, 6, 5, 5));
        for (int i = 0; i < SIZE; i++)
        {
            final int counter = i;
            buttons[i] = new JButton();
            buttons[i].setActionCommand("" + i);
            buttons[i].addActionListener(new ActionListener()
            {
                public void actionPerformed(ActionEvent ae)
                {
                    desiredPos = Integer.parseInt(
                                    (String) buttons[counter].getActionCommand());
                    timer.start();              
                }
            });
            contentPane.add(buttons[i]);
        }
        buttons[presentPos].setIcon(infoIcon);

        frame.setContentPane(contentPane);
        frame.pack();
        frame.setLocationByPlatform(true);
        frame.setVisible(true);
        timer = new Timer(1000, timerAction);
    }

    public static void main(String... args)
    {
        SwingUtilities.invokeLater(new Runnable()
        {
            public void run()
            {
                new GridExample().createAndDisplayGUI();
            }
        });
    }
}

Answer 2

这是因为在UI线程中执行do {}。要解决这个问题，您应该使用SwingWorker或javax.swing.Timer