Question

所以我一直在研究这个Java程序，计算机询问用户是否想知道数字的平方。如果用户的答案是“Y”或“y”，则计算机向用户询问他想要知道的数字，并打印数字的平方。然后计算机询问用户他是否想知道另一个号码的平方。如果答案是'Y'或'y'，计算机将获取一个数字并打印正方形。再次询问用户是否想知道另一个号码的平方。像这样，它继续下去。

我还为用户的答案是否定的，或既不是正面也不是负面的情况编写代码。通过这个，我的意思是用户的答案是'N'或'n'。这个程序在某个地方遇到问题，并且不能按照我想要的方式从A到Z工作。看，当我运行代码时，计算机会问我是否想知道数字的平方，然后输入'y'。然后我被问到我想知道的数字。我输入数字，计算机打印出它的正方形。然后我被问到我是否想知道另一个数字的平方。我输入'y'。然后我被问到我想知道的数字，我输入一个数字。计算机打印数字的平方。然后程序询问我是否想要知道另一个数字的平方并且刚刚结束，而它应该采取我的答案。我已经工作了近16到17个小时试图找到这个bug，但我做不到。你能不能测试代码并告诉我哪里出错了？感谢。

这是代码：

package looppracticea;

import java.util.Scanner;

public class LoopPracticeA {

    public static void main(String[] args) {

        Scanner geek = new Scanner(System.in);
        Scanner geek2 = new Scanner(System.in);

        long taken_number, taken_number2;
        String answer, answer2, answer_not_recognized, if_answer_is_no;
        answer_not_recognized = "You didn't enter any of the recognized answers.";
        if_answer_is_no = "'Kay. Whatever.";

        System.out.println("Do you want to know the square of a number? (Y/N)");
        answer = geek.nextLine();

        if (answer.equals("Y") || answer.equals("y")) {
            System.out.println("The number you want to know the square of is:");
            taken_number = geek.nextLong();
            System.out.println("That number squared is " + taken_number * taken_number + ".");
            System.out.println("Do you want to know the square of another number? (Y/N)");
            answer2 = geek2.nextLine();

            switch (answer2) {
                case "N":
                    System.out.println(if_answer_is_no);
                    break;
                case "n":
                    System.out.println(if_answer_is_no);
                    break;

                case "Y":
                    while (answer2.equals("Y")) {
                           System.out.println("The number you want to know the square of is:");
                           taken_number2 = geek2.nextLong();
                           System.out.println("That number squared is " + taken_number2 * taken_number2 + ".");
                           System.out.println("Do you want to know the square of another number? (Y/N)");
                           answer2 = geek2.nextLine();
                    }
                break;

                case "y":
                    while (answer2.equals("y")) {
                           System.out.println("The number you want to know the square of is:");
                           taken_number2 = geek2.nextLong();
                           System.out.println("That number squared is " + taken_number2 * taken_number2 + ".");
                           System.out.println("Do you want to know the square of another number? (Y/N)");
                           answer2 = geek2.nextLine();
                    }
                break;

                default:
                    System.out.println(answer_not_recognized);
                    break;
            }


        } else if (answer.equals("N") || answer.equals("n")) {
            System.out.println(if_answer_is_no);
        } else {
            System.out.println();
            System.out.println(answer_not_recognized);
        }

    }

}

Answer 1

以下是您的主要内容的固定内容：

        Scanner geek = new Scanner(System.in);

        long taken_number;
        String answer;
        String answer_not_recognized = "You didn't enter any of the recognized answers.";
        String if_answer_is_no = "'Kay. Whatever.";
        boolean firstTime = true;

        while(true)
        {
            if(firstTime)
                System.out.println("Do you want to know the square of a number? (Y/N)");
            else
                System.out.println("Do you want to know the square of another number? (Y/N)");
            answer = geek.nextLine();
            if (answer.equalsIgnoreCase("y"))
            {
                System.out.println("The number you want to know the square of is:");
                taken_number = geek.nextLong();
                System.out.println("That number squared is " + taken_number * taken_number + ".");
                answer = geek.nextLine();
            }
            else if (answer.equalsIgnoreCase("n")) {
                System.out.println(if_answer_is_no);
                break;
            } else {
                System.out.println();
                System.out.println(answer_not_recognized);
                break;
            }
            if(firstTime)
                firstTime = false;
        }

问题是您必须两次读取该行才能到达您想要的行。
第一个nextLine将读取输入

我还通过以下方式减少了代码：

只留下一个Scanner，因为这足够了
使用equalsIgnoreCase代替2 equals
只使用了一个answer和taken_number变量，因为这足够了
将变量声明放在不同的行上，因为它更容易维护约定。
将两个输入读数合并为一个，用布尔告诉我这是否是第一次，所以我知道要输出什么信息。
读取nextLine() 2次，因为第一个是输入

无法评估用户输入

1 个答案: