java - 如何提示用户输入 0 到 100 之间的数字？

Question

尝试将用户输入作为 0 到 100 之间的整数，并提示用户“再试一次”，只要他们的输入不符合此条件。

到目前为止，我的代码在一定程度上成功地实现了我的目标:

import java.util.Scanner;

public class FinalTestGrade
{
    public static void main(String[] args)
    {
        Scanner studentInput = new Scanner (System.in );
        int testGrade;

        System.out.println("Please enter your test grade (0 to 100)");
        while (!studentInput.hasNextInt()) 
            {
            System.out.println("Your input does not match the criteria, please enter a number between 0 and 100");
            studentInput.next();
            }
        testGrade = studentInput.nextInt();

        while (testGrade > 100 || testGrade < 0) 
            {
            System.out.println("Your input does not match the criteria, please enter a number between 0 and 100");
            testGrade = studentInput.nextInt();
            }

如您所见，程序将检查输入是否为 int。一旦用户成功输入一个 int，程序检查以确保他们的输入在 0 到 100 之间。当用户输入一个非 int 作为响应时，问题就出现了 到第二个提示(由第二个 while 循环启动)。下面是一个例子:

run:
Please enter your test grade (0 to 100)
P
Your input does not match the criteria, please enter a number between 0 and 100
109
Your input does not match the criteria, please enter a number between 0 and 100
P
    Exception in thread "main" java.util.InputMismatchException
            at java.util.Scanner.throwFor(Scanner.java:840)
            at java.util.Scanner.next(Scanner.java:1461)
            at java.util.Scanner.nextInt(Scanner.java:2091)
            at java.util.Scanner.nextInt(Scanner.java:2050)
            at finaltestgrade.FinalTestGrade.main(FinalTestGrade.java:24)
Java Result: 1
BUILD SUCCESSFUL (total time: 9 seconds)

长话短说，我想知道是否有一种方法可以组合我的 while 循环，以便接受表示为 0 到 100 之间的 int 的输入并将其保存为变量。所有不符合此标准的输入都应触发重复提示，直到输入符合此标准。有什么建议吗？

Answer 1

int testGrade = -1 ;
Scanner studentInput = new Scanner(System.in);
while (testGrade > 100 || testGrade < 0) 
{
   System.out.println("Your input does not match the criteria, please enter a number between 0 and 100");

   while(!studentInput.hasNextInt())
   {
       studentInput.next() ;
   }
   testGrade = studentInput.nextInt();
}

有一个无限循环来检查流中是否有无效字符。如果是，则使用它，这就是 hasNextInt() 的用途。如果您输入有效内容，该循环就会退出。