如何在一个 ActionListener 中有两个依赖的动作？

Question

我正在尝试在 java 中编写带有 GUI 的西洋跳棋 AI 程序。到目前为止，我已经设法初始化并填充了板（现在写为“B”和“W”）。我已经为开发板创建了一个 2D JButton 面板。

搬棋子的时候不知道怎么操作。我当前的问题是我需要玩家 select 一个预先存在的棋子（动作 1），然后一个空的 space 来放置 selected 棋子（动作 2）。当然，这些操作是相互依赖的，我需要先执行操作 1。

这是我目前得到的：

import javax.swing.*;
import java.awt.*;
import java.awt.event.ActionEvent;
import java.awt.event.ActionListener;
import java.util.Random;

public class Checkers implements ActionListener{

        private static final int BOARD_SIZE = 8;

        Random random = new Random();
        JFrame frame = new JFrame();
        JPanel title_panel = new JPanel();
        JPanel button_panel = new JPanel();
        JLabel textfield = new JLabel();
        JButton[][] buttons = new JButton[BOARD_SIZE][BOARD_SIZE];

        Checkers(){
            frame.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
            frame.setSize(800,800);
            frame.getContentPane().setBackground(new Color(50, 50, 50));
            frame.setLayout(new BorderLayout());
            frame.setVisible(true);

            textfield.setBackground(new Color(25,25,25));
            textfield.setForeground(new Color(25, 255, 0));
            textfield.setFont(new Font("Arial", Font.BOLD, 75));
            textfield.setHorizontalAlignment(JLabel.CENTER);
            textfield.setText("Checkers");
            textfield.setOpaque(true);

            title_panel.setLayout(new BorderLayout());
            title_panel.setBounds(0,0,800, 100);
            button_panel.setLayout(new GridLayout(BOARD_SIZE,BOARD_SIZE));
            button_panel.setBackground(new Color(150,150,150));

            for (int i=0; i<BOARD_SIZE; i++){
                for (int j=0; j<BOARD_SIZE; j++){
                    buttons[i][j] = new JButton();
                    button_panel.add(buttons[i][j]);
                    buttons[i][j].setFont(new Font("Arial",Font.BOLD, 30));
                    buttons[i][j].setFocusable(false);
                    buttons[i][j].addActionListener(this);
                }
            }

            title_panel.add(textfield);
            frame.add(title_panel, BorderLayout.NORTH);
            frame.add(button_panel);
            initialiseBoard(buttons);
        }

    @Override
    public void actionPerformed(ActionEvent e) {
        for (int i = 0; i < BOARD_SIZE; i++) {
            for (int j = 0; j < BOARD_SIZE; j++) {
                if (e.getSource() == buttons[i][j]) {
                    String moving_piece = buttons[i][j].getText();
                }
                }
            }
        }

    // method that fills the initial board.
    public void initialiseBoard(JButton[][] buttons){
        for (int i=0; i<BOARD_SIZE; i++){
            for (int j=0; j<BOARD_SIZE; j++){

                if (i<3){
                    if (i%2==0){
                        if (j%2 != 0){
                            buttons[i][j].setText("W");
                        }
                    }
                    else {
                        if (j%2==0){
                            buttons[i][j].setText("W");
                        }
                    }
                }
                if (i>=BOARD_SIZE-3){
                    if (i%2 == 0){
                        if (j%2 != 0){
                            buttons[i][j].setText("B");
                        }
                    }
                    else {
                        if (j%2==0){
                            buttons[i][j].setText("B");
                        }
                    }
                }
            }
        }
    }
}

Answer 1

你需要记住和管理一个片段的选择。您可以通过以下方式完成：

//Introduce a field to store selected button
private JButton selected = null ;

并修改ActionListener：

@Override
public void actionPerformed(ActionEvent e) {

    JButton clicked = (JButton) e.getSource();
    if(selected != null ){
        clicked.setText(selected.getText());
        selected.setText("");
        selected = null;
    }else{
        if( ! clicked.getText().isEmpty()) {
            selected = clicked;
        }
    }
}

考虑使用 JToggleButtons 而不是 JButtons 。

Answer 2

正如我在评论中提到的，解决方案的关键是让程序的侦听器基于“state-dependent 行为”，这意味着侦听器的行为将取决于程序的状态。

例如，如果您为您的程序提供一个实例字段，例如 selectedButton，它包含对最后选择的按钮的引用，那么任何 ActionListener 的响应都可能因该字段包含的内容而异。第一次按下任何 JButton 时，此字段将保持 null，因为之前没有按下 JButton，因此 ActionListener 会将此按钮分配给该字段。第二次调用 ActionListener（第二次按下 JButton）时，该字段将保存一个值，因此，使用 if / else 块检查该字段的状态时，侦听器的行为可能会有所不同：

public class Checkers2 {
    private static final int BOARD_SIZE = 8;
    // .....
    private JButton[][] buttons = new JButton[BOARD_SIZE][BOARD_SIZE];
    private JButton selectedButton = null;
    
    public Checkers2() {
        buttonPanel.setLayout(new GridLayout(BOARD_SIZE, BOARD_SIZE));
        for (int i = 0; i < buttons.length; i++) {
            for (int j = 0; j < buttons[i].length; j++) {
                buttons[i][j] = new JButton(" ");
                buttonPanel.add(buttons[i][j]);
                buttons[i][j].setFont(BTN_FONT);
                buttons[i][j].setPreferredSize(BTN_SIZE);
                buttons[i][j].addActionListener(e -> btnsActionPerormed(e));
            }
        }
        
        // ...
    }
    
    private void btnsActionPerormed(ActionEvent e) {
        JButton source = (JButton) e.getSource();
        
        // if selectedButton is null, then this is the first press      
        if (selectedButton == null) {
            if (source.getText().trim().isEmpty()) {
                return; // no button text means that there is no piece here. Exit method
            } else {
                // first set the selectedButton's value
                selectedButton = source;
                
                // change its appearance so we know that it has been pressed
                source.setForeground(Color.RED);
                source.setBackground(Color.LIGHT_GRAY);
            }
        } else {
            
            // else, if selectedButton is NOT null, then this is the second press
            // only do things if this current button is empty, if it has no text
            if (source.getText().trim().isEmpty()) {
                selectedButton.setForeground(Color.BLACK);
                selectedButton.setBackground(null);
                source.setText(selectedButton.getText());
                selectedButton.setText("");
                
                // the code below is of key importance
                selectedButton = null;
            }
        }
    }

显示的最后一位代码也是关键——在第二次按下 JButton 时，我们 re-set selectedButton 返回到 null，这样程序的状态就被设置回它的初始状态，并且监听器的初始行为可以重复。

演示这一点的简单示例程序如下所示。该程序仍未合并真正的 Checker 游戏逻辑，程序在该逻辑中确定移动是否真正合法：

import java.awt.Color;
import java.awt.Dimension;
import java.awt.Font;
import java.awt.GridLayout;
import java.awt.event.ActionEvent;
import javax.swing.*;

public class Checkers2 {
    private static final int BOARD_SIZE = 8;
    private static final Font BTN_FONT = new Font("Arial", Font.BOLD, 30);
    private static final Dimension BTN_SIZE = new Dimension(100, 100);
    private JPanel buttonPanel = new JPanel();
    private JButton[][] buttons = new JButton[BOARD_SIZE][BOARD_SIZE];
    private JButton selectedButton = null;

    public Checkers2() {
        buttonPanel.setLayout(new GridLayout(BOARD_SIZE, BOARD_SIZE));
        for (int i = 0; i < buttons.length; i++) {
            for (int j = 0; j < buttons[i].length; j++) {
                buttons[i][j] = new JButton(" ");
                buttonPanel.add(buttons[i][j]);
                buttons[i][j].setFont(BTN_FONT);
                buttons[i][j].setPreferredSize(BTN_SIZE);
                buttons[i][j].addActionListener(e -> btnsActionPerormed(e));
            }
        }

        // Place "W" and "B" text in appropriate locations:
        for (int i = 0; i < 12; i++) {
            int wI = (2 * i) / BOARD_SIZE;
            int wJ = (2 * i) % BOARD_SIZE + ((wI % 2 == 1) ? 0 : 1);

            buttons[wI][wJ].setText("W");
            buttons[BOARD_SIZE - wI - 1][BOARD_SIZE - wJ - 1].setText("B");
        }
    }

    private void btnsActionPerormed(ActionEvent e) {
        JButton source = (JButton) e.getSource();

        // if selectedButton is null, then this is the first press
        if (selectedButton == null) {
            if (source.getText().trim().isEmpty()) {
                return; // no button text means that there is no piece here. Exit method
            } else {
                // first set the selectedButton's value
                selectedButton = source;

                // change its appearance so we know that it has been pressed
                source.setForeground(Color.RED);
                source.setBackground(Color.LIGHT_GRAY);
            }
        } else {

            // else, if selectedButton is NOT null, then this is the second press
            // only do things if this current button is empty, if it has no text
            if (source.getText().trim().isEmpty()) {
                selectedButton.setForeground(Color.BLACK);
                selectedButton.setBackground(null);
                source.setText(selectedButton.getText());
                selectedButton.setText("");

                // the code below is of key importance
                selectedButton = null;
            }
        }
    }

    public JPanel getButtonPanel() {
        return buttonPanel;
    }

    public static void main(String[] args) {
        SwingUtilities.invokeLater(() -> {
            Checkers2 checkers = new Checkers2();

            JFrame frame = new JFrame("Checkers");
            frame.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
            frame.add(checkers.getButtonPanel());
            frame.pack();
            frame.setLocationRelativeTo(null);
            frame.setVisible(true);
        });
    }

}

并且我希望展示如何在编辑中合并程序逻辑。为此，我建议使用“Model-View-Controller”或 MVC 程序结构。

如何在一个 ActionListener 中有两个依赖的动作？

How to have two dependent actions within an ActionListener?

java

swing

jframe