Question

我正试图找一种方法向用户询问是/否问题，该问题验证输入仅为是（y / Y）或否（n / N）。我试图减少变量，同时仍然保持可读性。到目前为止，这是我提出的：

import java.util.Scanner;

public class InputValidation
{
    public static void main(String[] args)
    {
        Scanner input = new Scanner(System.in);
        String answer;

        do
        {
            System.out.print("Do you wish to proceed? (Y/N): ");
            answer = input.next().trim().toUpperCase();
            yesAnswer = answer.equals("Y");
        } while (answer.matches("[YN]"));

        if (yesAnswer)
        {
            // Do something
        }
        else
        {
            // Do something else
        }
    }
}

我不一定要寻找最少量的角色;更多简短而灵活的东西。有什么想法吗？

Answer 1

根据您想要实现的目标，我可能只是在确定实际结果之前直接评估输入，例如

do {
    System.out.print("Do you wish to proceed? (Y/N): ");
    answer = input.next().trim().toUpperCase();
} while (!answer.matches("[YN]"));
boolean yesAnswer = answer.equalsIgnoreCase("Y");

在确定answer之前，这基本上等到Y匹配N，y，n，yesAnswer。我接下来要做的就是将它包装在方法中......

if (askYesNo("Do you wish to proceed? (Y/N): ")) {...

然后你可以继续为方法提供可选的退出条件..

askYesNo("Delete all files [Ok/Cancel]", "(ok)?", "(cancel)?")

例如......

public class YesNo {

    public static void main(String[] args) {
        if (askYesNo("Do you wish to proceed? (Y/N): ")) {
            System.out.println("Okay...");
        }

        if (askYesNo("Delete all Files? (Okay/Cancel): ", "okay", "cancel")) {
            System.out.println("Make it so");
        }
    }

    public static boolean askYesNo(String question) {
        return askYesNo(question, "[Y]", "[N]");
    }

    public static boolean askYesNo(String question, String positive, String negative) {
        Scanner input = new Scanner(System.in);
        // Convert everything to upper case for simplicity...
        positive = postive.toUpperCase();
        negative = negative.toUpperCase();
        String answer;
        do {
            System.out.print(question);
            answer = input.next().trim().toUpperCase();
        } while (!answer.matches(positive) && !answer.matches(negative));
        // Assess if we match a positive response
        return answer.matches(positive);
    }
}

要求确认Java的简短方法

1 个答案: