Java JOptionPane 需要帮助从错误后留下的位置启动程序

Question

在下面的程序中，我创建了一个简单的数字游戏，选择 1 到 100 之间的数字并要求用户猜测该数字。用户可以继续猜测，直到尝试了 10 次为止。

当我运行程序并输入字符串时，程序抛出 NumberFormatException。单击“确定”后，我找不到处理此问题的方法，因此我调用了“TryAgain”方法。我想知道的是，在我不小心按下带有字符串文本的按钮后，如何以相同的尝试次数重新启动程序。

import javax.swing.JOptionPane;

public class NumberGame {
    static int scoreOfWins = 0; 
    static int numberOfAttempt = 0;
    public static void main(String[] args) {
        Game();
    }

    static void Game() {
        try {
            int num = (int) (Math.random() * 100);
            int count = 10;
            while(count > 0) {
//              System.out.println(num);
                String inputTitle = "Guess the number between 1 to 100";
                String input = JOptionPane.showInputDialog(inputTitle, "Enter your answer here...");
                if (input == null) {
                    System.exit(0);
                }
                int intInput = Integer.parseInt(input);
                System.out.println(intInput);
                if (intInput == num) {
                    numberOfAttempt = 11 - count;
                    JOptionPane.showMessageDialog(null, "You win! Your guess was right after " + numberOfAttempt + " attempts");
                    scoreOfWins++;
                    TryAgain();
                    return;
                } else if (intInput > num) {
                    String numberOfTimes = " You can try " + (count - 1) + " times";
                    JOptionPane.showMessageDialog(null, "Your guess was not right! Please try Again with a lower value." + numberOfTimes);
                    count--;
                } else if (intInput < num){
                    String numberOfTimes = " You can try " + (count - 1) + " times";
                    JOptionPane.showMessageDialog(null, "Your guess was not right! Please try Again with a higher value." + numberOfTimes);
                    count--;
                }
            }
            JOptionPane.showMessageDialog(null, "You Lost the game!");
            TryAgain();
        } catch (Exception E) {
            JOptionPane.showMessageDialog(null, "Number format exception detected, Please only input numbers!");
            TryAgain();
        }
    }
    static void TryAgain() {
        while (true) {
            String inputAns = JOptionPane.showInputDialog("Type yes if you want to containue!"
                    + " Type No if you want to give up!", "you won " + scoreOfWins + " times do you want to continue?");
            if (inputAns == null) {
                System.exit(0);
            }
            inputAns = inputAns.toUpperCase();
            if (inputAns.equals("Y") | inputAns.equals("Yes")) {
                Game();
            } else if (inputAns.equals("N") |inputAns.equals("NO")) {
                JOptionPane.showMessageDialog(null, "OKAY you won " + scoreOfWins + " times today, see you later!");
                return;
            } else {
                JOptionPane.showMessageDialog(null, "Please provide a good valid identifier like Yes or NO");
            }
        }
    }
}

Answer 1

这是您的代码，正在执行您想要的操作（可能不是最有效的方法）

public class NumberGame {

    static int scoreOfWins = 0; 
    static int numberOfAttempt = 0;
    
    public static void main(String[] args) {
        game();
    }
    
    static int num = new Random().nextInt(100) + 1;
    static int count = 10;

    static void game() {
        try {
            while(count > 0) {
                String inputTitle = "Guess the number between 1 to 100";
                String input = JOptionPane.showInputDialog(inputTitle, "Enter your answer here...");
                if (input == null) {
                    System.exit(0);
                }
                int intInput = Integer.parseInt(input);
                System.out.println(intInput);
                if (intInput == num) {
                    numberOfAttempt = 11 - count;
                    JOptionPane.showMessageDialog(null, "You win! Your guess was right after " + numberOfAttempt + " attempts");
                    scoreOfWins++;
                    tryAgain();
                    return;
                } else if (intInput > num) {
                    String numberOfTimes = " You can try " + (count - 1) + " times";
                    JOptionPane.showMessageDialog(null, "Your guess was not right! Please try Again with a lower value." + numberOfTimes);
                    count--;
                } else if (intInput < num){
                    String numberOfTimes = " You can try " + (count - 1) + " times";
                    JOptionPane.showMessageDialog(null, "Your guess was not right! Please try Again with a higher value." + numberOfTimes);
                    count--;
                }
            }
            JOptionPane.showMessageDialog(null, "You Lost the game!");
            tryAgain();
        } catch (Exception E) {
            JOptionPane.showMessageDialog(null, "Number format exception detected, Please only input numbers!");
            game();
        }
    }
    
    static void restart() {
        count = 0;
        game();
    }
    
    static void tryAgain() {
        while (true) {
            String inputAns = JOptionPane.showInputDialog("Type yes if you want to containue!"
                    + " Type No if you want to give up!", "you won " + scoreOfWins + " times do you want to continue?");
            if (inputAns == null) {
                System.exit(0);
            }
            inputAns = inputAns.toUpperCase();
            if (inputAns.equals("Y") | inputAns.equalsIgnoreCase("yes")) {
                restart();
            } else if (inputAns.equals("N") |inputAns.equalsIgnoreCase("no")) {
                JOptionPane.showMessageDialog(null, "OKAY you won " + scoreOfWins + " times today, see you later!");
                return;
            } else {
                JOptionPane.showMessageDialog(null, "Please provide a good valid identifier like Yes or NO");
            }
        }
    }
}

所以我在你的代码中修复的第一件事是一个小但很好的练习，那就是命名约定。这是非常不言自明的。

其次，我使用了

inputAns.equalsIgnoreCase("yes")

和

inputAns.equalsIgnoreCase("no")

，它允许用户以他们想要的任何方式输入“yes”，甚至是“Yes”。

接下来我实现了

new Random().nextInt(100) + 1

，这可能是生成随机数的更好方法。

除了从游戏方法中删除

num

和

count

并将其设为一个字段之外，我通常没有更改您的很多代码。现在，每次运行游戏方法时，计数都不会改变，只有当您重新启动游戏时才会改变。