java - 当我从cmd运行jar时发生InputMismatchException，如果从想法运行则不会发生

在我的主要方法中，我有以下代码片段:

            try {
                select = scanner.nextInt();
            } catch (InputMismatchException e) {
                scanner.next(); //we should read erroneous
                System.out.println("Error. Please input number.");
                continue;
            }

我让mvn clezan install->转到目标目录。并开始使用应用程序
实际上，我输入了数字，但是在控制台中，我看到以下消息:

D:\freelance\Новая папка\myrepository\target>java -jar palindrome-artifactId-1.0
-SNAPSHOT.jar
Please type your name:
u1
Please select menu item
1 - suggest word, 2 - change user, 3 - my score, 4 - my word list, 5 - records,
6 - exit
1
Error. Please input number.
Please select menu item
1 - suggest word, 2 - change user, 3 - my score, 4 - my word list, 5 - records,
6 - exit

当我从idea(选择main方法)调用应用程序时，我没有看到这个问题。

请帮助解决我的问题。

附言

完整的主要方法:

public static void main(String[] args) {
    System.out.println("Please type your name:");
    try (Scanner scanner = new Scanner(System.in)) {
        scanner.useDelimiter("\n");
        String userName = scanner.next();
        Game game = new Game(userName);
        AtomicInteger atomicInteger = new AtomicInteger();
        int select = 0;
        do {
            System.out.println("Please select menu item");
            System.out.println("1 - suggest word, 2 - change user, 3 - my score, 4 - my word list, 5 - records, 6 - exit");
            try {
                select = scanner.nextInt();
            } catch (InputMismatchException e) {
                scanner.next(); //we should read erroneous
                System.out.println("Error. Please input number.");
                continue;
            }

            switch (select) {
                case 1:
                    System.out.print("Word:");
                    String word = scanner.next();
                    if (game.suggestWord(word)) {
                        System.out.println("Accepted: your score - " + game.getCurrentUserScore());
                    } else {
                        System.out.println("Rejected: Word already exists in your list or it is not palindrome");
                        System.out.println("Your score - " + game.getCurrentUserScore());
                    }
                    break;
                case 2:
                    System.out.print("Name:");
                    String name = scanner.next();
                    game.changeUser(name);
                    System.out.println("User changed successfully. Your score - " + game.getCurrentUserScore());
                    break;
                case 3:
                    System.out.println("Your score - " + game.getCurrentUserScore());
                    break;
                case 4:
                    System.out.println("Accepted words:");
                    game.getCurrentUserAcceptedWords().forEach(System.out::println);
                    break;
                case 5:
                    atomicInteger.set(1);
                    game.getScores().forEach((k, v) -> System.out.println("#" + atomicInteger.getAndIncrement() + ". name: " + k + " , score: " + v));
                    break;
                case 6:
                    System.out.println("Goodbye! thanks for the game");
                    break;
                default:
                    System.out.println("You selected nonexistent menu item. Please try one more time.");
            }

        } while (select != 6);
    }
}

最佳答案

您的问题可能是定界符。它正在寻找\n，但是在Windows上，每行以\r\n结尾。因此，尽管您没有看到"u1\r"(检查其长度)，但您获得的用户名可能是\r，并且您输入的数字被读取为"1\r"，该数字不能解析为数字。

我想您的IDE中的模拟控制台会将您的紧迫返回解释为\n，这就是它在那里起作用的原因。

因此，将您的分隔符(而不是\n)更改为\r?\n。分隔符是一个正则表达式，这意味着“可选的回车符，后跟换行符”。