Java猜数字游戏异常后如何从当前剩余尝试次数恢复运行

我编写了一个简单的猜数字游戏：程序随机生成1到100之间的数字，用户最多可进行10次猜测尝试。当前程序存在一个问题：当用户输入字符串时会抛出NumberFormatException，点击弹窗的“OK”后，目前只能调用TryAgain方法重新开始游戏，但我希望实现从当前剩余的尝试次数处继续运行程序，而非重新开始。

原代码 #

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");
            }
        }
    }
}

解决方案 #

问题核心是你把整个游戏逻辑包裹在了外层try-catch中，触发异常后直接跳出循环并重启游戏。要实现从剩余次数继续，需将异常捕获范围缩小到仅输入解析环节，确保异常不会中断整个循环：

1. 将Integer.parseInt(input)单独放入try-catch块
2. 捕获到NumberFormatException时，仅提示错误，用continue跳过当前循环，保留剩余尝试次数

修改后的代码 #

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() {
        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);
            }
            
            // 单独捕获输入解析异常
            int intInput;
            try {
                intInput = Integer.parseInt(input);
            } catch (NumberFormatException e) {
                JOptionPane.showMessageDialog(null, "Number format exception detected, Please only input numbers!");
                continue; // 跳过当前循环，次数不减少
            }
            
            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();
    }
    
    static void TryAgain() {
        while (true) {
            String inputAns = JOptionPane.showInputDialog("Type yes if you want to continue!"
                    + " 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();
            // 替换为短路或运算符，修正拼写错误
            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 valid input like Yes or NO");
            }
        }
    }
}

额外优化：

• 将|替换为短路或||，符合Java常规写法
• 修正TryAgain方法中的拼写错误containue为continue
• 捕获具体的NumberFormatException而非泛型Exception，避免误捕无关异常

内容的提问来源于stack exchange，提问作者Wahid Ahmad Rustaqi