当我点击 n 或 N 时程序没有退出？

Question

我有以下作业问题：

Q1. Use nested for loops statements to draw empty boxes of any character (input from user). The boxes have the same number of rows and columns (input from the user; valid range: 5 to 21). Test for errors in input (including type)

SAMPLE OUTPUT:
Do you want to start(Y/N): y
How many chars/last row? n
Not an integer! Try again! How many chars/last row? fgfgfg
Not an integer! Try again! How many chars/last row? 7.6
Not an integer! Try again! How many chars/last row? 34
ERROR! Valid range 5 - 21. How many chars/last row? 7
What character? k

Do you want to continue(Y/N): y

我写了下面的代码，但是当我点击 'n' 或 'N' 时它没有退出，我不确定为什么。我将如何解决这个问题？

public static void main(String[] args) {
    Scanner input = new Scanner(System. in );
    char answer = 'n';
    int row = 0;
    char output = 'k';

    do {
        System.out.println("DO YOU WANT TO START Y OR N?");
        answer = input.next().charAt(0);
        System.out.println("enter the number of rows");

        while (!input.hasNextInt()) {
            System.out.println("Not an integer,try again ");
            input.next();
        }

        row = input.nextInt();


        while (row < 5 || row > 21) {
            System.out.println("ERROR! Valid range 5 - 21. How many chars/last row?");
            row = input.nextInt();
        }
        System.out.println("WHAT CHARACTER?");
        output = input.next().charAt(0);

        for (int i = 0; i < row; i++) { //nested for loop to create the box
            System.out.print(output);
        }
        System.out.println();

        for (int i = 0; i < row - 2; i++) {
            System.out.print(output);
            for (int j = 0; j < row - 2; j++) {
                System.out.print(" ");
            }
            System.out.print(output);
            System.out.println();
        }
        for (int i = 0; i < row; i++) {
            System.out.print(output);
        }
        System.out.println();
        System.out.println();
        System.out.println("DO YOU WANT TO CONTINUE ? Y OR N");
        answer = input.next().charAt(0);

    } while (answer == 'Y' || answer == 'y');

    input.close();
    System.out.println("game stop");
}

Answer 1

最简单的方法是：

检查循环内的输入 "N"，然后跳出 while 循环，可能像这样：

if ((answer == 'n') || (answer == 'N')) {
    break;
}

此外，您正在该程序中检查 y/n 输入 2 次。一个更好的写法是使用普通的 while 循环而不是 do-while 循环；显然，在这个问题中，如果您在一开始就输入 N，那么您根本不应该运行ning 通过该程序。 Do-While 循环可用于确保程序运行至少执行一次（这不是此处应发生的情况；如果输入有效，程序应仅运行例如："y").虽然使用 DW 循环是 "ok"，但 while 循环在这里会更好。

你的循环可以这样写：

// you need this line to initially print the y/n question.
System.out.println("DO YOU WANT TO START Y OR N?");
// get an input and check if it is not n
while (Character.toUpperCase(input.next().charAt(0)) != 'N') {    
    // do stuff here 
    System.out.println("DO YOU WANT TO CONTINUE ? Y OR N"); // ask for next input
}

Answer 2

您需要在 Do you want to start(Y/N): 和 Do you want to continue(Y/N):

之后添加 N 的条件

System.exit(0)用于终止程序。

输入这段代码

System.out.println("DO YOU WANT TO START Y OR N?");
    answer = input.next().charAt(0);
    if(answer == n || answer == N){
        System.exit(0);
    }

这是 Do you want to continue(Y/N):

System.out.println("DO YOU WANT TO CONTINUE ? Y OR N");
    answer = input.next().charAt(0);
    if(answer == n || answer == N){
        System.exit(0);
    }

编辑

如果你想打印'Game Stop'如果答案是N，那么在System.exit(0)

之前使用Thread.sleep(timeInMilliseconds);

if(answer == n || answer == N){
    Thread.sleep(5000); //This will make console wait for 5 seconds before exiting.
    System.out.println("Game Stop."); //game stop will be printed for 5 seconds
    System.exit(0);
}

当我点击 n 或 N 时程序没有退出？

Program doesn't exit when I hit n or N?

java

loops

while-loop

sentinel