Question

嘿伙计们我正在尝试制作匹配卡片的游戏。如果两张牌匹配，用户获得一个点并且卡片保持可见，否则翻转它们（或setText（＆＃34;＆＃34;））我做了关于摇摆睡眠的研究，但我不确定如何实现它我的代码。我尝试了一切，但我不能让它发挥作用。我在main中运行此代码。

ActionListener buttonListener = new ActionListener() {

        @Override
        public void actionPerformed(ActionEvent e)
        {
            JButton selectedButton = (JButton)e.getSource();

            for (int row = 0; row < 6;row++){
                    for(int col = 0; col < 6; col++){
                        if (buttons[row][col] == selectedButton){
                            flipCard(row, col);
                            if(stack.empty()){
                                stack.push(row+","+col);
                            }else{
                                String word = (String)stack.pop();
                                String[] ar = word.split(",");
                                System.out.println(ar[0] + " " + ar[1]);
                                if (cardList.getCardNode(row, col).getLetter() ==
                                        cardList.getCardNode(Integer.parseInt(ar[0]),
                                        Integer.parseInt(ar[1])).getLetter()){
                                    System.out.println("equal");
                                }else{
                                    System.out.println("not equal");
                                    //Compiler complains 
                                    //Exception in thread "AWT-EventQueue-0" java.lang.ClassCastException: javax.swing.Timer cannot be cast to javax.swing.JButton
                                    Timer timer = new Timer(100 ,this);
                                    timer.setRepeats(false);
                                    timer.start();
                                    buttons[row][col].setText("");
                                    buttons[Integer.parseInt(ar[0])]
                                            [Integer.parseInt(ar[1])].setText("");
                                }
                            }
                        }
                    }
            }
        }
    };

Answer 1

我“想”你应该做的更像是......

Timer timer = new Timer(100, new ActionListener() {
    @Override
    public void actionPerformed(ActionEvent evt) {
        //Pop stack coordinates and set them back to ""
        //setText on button clicked to ""
        System.out.println(cardList.getCardNode(row, col).getLetter());        
    }
});
timer.setRepeats(false);
timer.start();

在延迟100毫秒之后，将调用ActionListener的{{1}}方法，允许您重置UI的状态...

问题是我在循环内部并且只能在单击时访问row和col

然后创建一个actionPerformed，其中包含所需信息并在调用ActionListener方法时对其进行操作...

actionPerformed

然后将其与public class FlipperHandler implements ActionListener { private JButton[] buttons; private int[] card1, card2; public FlipperHandler(JButton[] buttons, int[] card1, int[] card2) { this.buttons = buttons; this.card1 = card1; this.card2 = card2; } @Override public void actionPerformed(ActionEvent evt) { buttons[card1[0]][card1[1]].setText(""); buttons[card2[0]][card2[2]].setText(""); } } ...

一起使用

Timer

如何睡一个按钮setText？

1 个答案: