Question

我有一个确认用户输入正确的验证类。此类中的一个方法尝试验证继续应用程序关键字。用户可以输入y或Y继续，n或N退出。任何其他字母都应该通过警告消息回答，然后是用户输入其答案的另一个机会。这是我为该方法编写的代码。

public static String getValidContinueCode(Scanner sc, String prompt)
    {
        String choice = "";
        boolean isValid = false;

        while (isValid == false)
        {
        System.out.print(prompt);

        if (sc.next().equalsIgnoreCase("y"))
        {
            choice = sc.next();
            isValid = true;
        }
        else if (sc.next().equalsIgnoreCase("n"))
        {
            choice = sc.next();
            isValid = true;
        }
        else
        {
            System.out.println("Please enter y or n");
        }
        }
        return choice;

    }

Answer 1

当您从具有sc.next()的扫描仪中读取该字符时，您无法再次阅读该字符。您需要在阅读时存储它并围绕读取值构建逻辑。

public static String getValidContinueCode(Scanner sc, String prompt) {
    String choice = "";
    boolean isValid = false;

    while (!isValid) {
        System.out.print(prompt);

        choice = sc.next();
        if (choice.equalsIgnoreCase("y")) {
            isValid = true;
        }
        else if (choice.equalsIgnoreCase("n")) {
            isValid = true;
        }
        else {
            System.out.println("Please enter y or n");
        }
    }
    return choice;

}

您可以做几项小改进。你可能只想返回一个带答案的布尔值。像这样。

public static boolean getValidContinueCode(Scanner sc, String prompt) {
    while (true) {
        System.out.print(prompt);

        if (sc.next().equalsIgnoreCase("y")) {
            return true;
        }
        else if (sc.next().equalsIgnoreCase("n")) {
            return false;
        }
        else {
            System.out.println("Please enter y or n");
        }
    }
}

Answer 2

试试这个

   public static void main(String[] args) {
    getValidContinueCode(new Scanner(System.in),"");
}

public static void getValidContinueCode(Scanner sc,String code) {
    code = "enter a letter";
    boolean status = true;
    while (status) {
        System.out.println(code);
        String check = sc.next();
        if (check.equalsIgnoreCase("y")) {
            sc = new Scanner(System.in);
            code = "ok";
            status = true;

        } else if (check.equalsIgnoreCase("n")) {
            sc = new Scanner(System.in);
            code = "exit";
            System.out.println("exit");
            status = false;
        } else {
            code = "retry";
            status = true;
        }

    }
}

验证用户输入以继续应用程序

2 个答案: