Question

我正在尝试仅将数字转换为0到100之间的单词，如果不满足条件而不退出该程序，如何重新执行该程序。如果我输入55，我想输出55，但是如果我键入的数字不在0-100之间，它将输出“您应该输入0-100仅重试一次”，然后它将自动转到“输入数字0之间仅限-100”

package convertnumbertowords;
import java.util.Scanner;
public class ConvertNumberToWords {
public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    System.out.println("Input number between 0-100 only ");
    int num1 = sc.nextInt();
    while (num1 <= -1 && num1 >= 101 ){
        if(num1 <= 100 && num1 >= 0){
            System.out.println("The "+num1+" in words is "+ 
            Integer.toString(num1));
        }
        else{
            System.out.println("You should input 0-100 only Try Again");
        }
    }
}
}

Answer 1

我的建议是将单词保留在一个数组中，将其分隔为一个，然后拉伸。例如：

String ones[] = { " ", " One", " Two", " Three", " Four", " Five", " Six", " Seven", " Eight", " Nine", " Ten"," Eleven", " Twelve", " Thirteen", " Fourteen", "Fifteen", "Sixteen", " Seventeen", " Eighteen"," Nineteen" };

我将ones [0]保留为空的原因，这样很容易理解，ones [1] =一个

String tens[] = { " ", " ", " Twenty", " Thirty", " Forty", " Fifty", " Sixty", "Seventy", " Eighty", " Ninety" };

我将0和10保留为空，因为我已经在其中声明了它。

根据我的示例，如果小于该值，则仅使用ones [number]。如果大于19，我建议将其除以10得到十个单词的单词，然后将数字模量（％）除以10得到一个单词。尽力而为，我的建议可能不是完美的，但希望它能给您一些想法

Answer 2

要保持输入正确的输入，您首先要输入->检查是否错误->如果是，则再次提示用户，否则继续下一步。在代码中看起来像这样

// Ask for input
System.out.println("Input number between 0-100 only ");
int num = sc.nextInt();
// Check if input is incorrect
while (num > 100 || num < 0) {
    // If input is incorrect prompt user again
    System.out.println("You should input 0-100 only Try Again");
    // Update num and check until unser enters a num within the range
    num = sc.nextInt();  
}

将数字转换为单词并在输入无效的情况下重新执行

2 个答案: