y/n输入验证，包括没有输入输入键

Question

我试图在一个方法中捕获任何输入(输入键)和无效输入( y/n除外)。我尝试了两种不同的方式(粘贴)，但我不能使工作同时“输入键”和“模糊类型/n”在一起。谢谢你的帮助。

第一次尝试：

public static String askToContinue(Scanner sc) {

String choice = "";
boolean isValid = false;

while (!isValid){System.out.print("Continue? (y/n): ");
    if (sc.hasNext()){
        choice = sc.next();
        isValid = true;
    } else {System.out.println("Error! "
                + "This entry is required. Try again");    
    }

    if (isValid && !choice.equals("y") || !choice.equals("n")) {
        System.out.println("Error! Entry must be 'y' or 'n'. Try again");
        isValid = false;
    }
   }
    //sc.nextLine();  // discard any other data entered on the line
    System.out.println();
    return choice;
   }        



   2nd attempt

    public static String askToContinue(Scanner sc) {
    System.out.print("Continue? (y/n): ");
    String choice;



    while (true) {choice = sc.next();

    //?????????????????????????????????????????????????????
        if (choice.length() == 0){ System.out.println("Error! "
                + "This entry is required. Try again");
            continue;
        }


        if (!(choice.equals("y") || choice.equals("n"))) {
            System.out.println("Error! Entry must be 'y' or 'n'. Try                   again");
            continue;
        }

        break;    
    }

    sc.nextLine();  // discard any other data entered on the line
    System.out.println();
    return choice;
    }

Answer 1

我第一次尝试了你的代码。我解释了注释行，它包含在下面的代码中，如；

public static String askToContinue(Scanner sc) {
    String choice = "";
    boolean isValid = false;
    while (!isValid) {
        System.out.print("Continue? (y/n): ");
        choice = sc.nextLine(); //to reads all line , because this cannot read with empty enter input
        isValid = true;
        if (choice.isEmpty()) {  //this isEmpty for empty enter
            System.out.println("Error! "
                    + "This entry is required. Try again");
        }
        System.out.println(choice);
        //this logic if not y or n , it will return error
        if (!choice.equals("y") && !choice.equals("n")) {
            System.out.println("Error! Entry must be 'y' or 'n'. Try again");
            isValid = false;
        }
    }
    //sc.nextLine();  // discard any other data entered on the line
    System.out.println();
    return choice;
}

Answer 2

在第一种情况下，您的if语句是错误的。您正在检查是否选择is not equal to 'y'、或 not equal to 'n'，这将始终是正确的。

变化

if (isValid && !choice.equals("y") || !choice.equals("n"))

至

if (isValid && !choice.equals("y") && !choice.equals("n"))