java - 在java中获取从 'try'到 'catch'的输入

Question

这是练习:

If the input is a valid grade number, the program should print "OK". Otherwise, the program should print the entered value and "is not a valid grade." and prompt a grade number again. The program should keep asking grade numbers until the user enters a valid grade number.

输出示例:

Enter grade (0-5): 9
9 is not a valid grade

Enter grade (0-5): two
two is not a valid grade

Enter grade (0-5): 2
OK

我在 while 循环中使用带有 NumberFormatException 的“try and catch”。但是，我无法像示例中的第二个那样打印输出。我的输出如下:

Enter grade (0-5): 9
9 is not a valid grade

Enter grade (0-5): two
9 is not a valid grade

Enter grade (0-5): 2
OK

新输入“two”未保存。我该如何解决这个问题？

下面是我的代码:

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

        System.out.print("Enter grade (0-5): ");
        int grade = Integer.parseInt(input.nextLine());


        if (0 <= grade && grade <= 5) {
            System.out.print("OK");
        } else {
            while (grade < 0 || grade > 5) {
                try {
                    System.out.println(grade + " is not a valid grade.");
                    System.out.print("\n");
                    System.out.print("Enter grade (0-5): ");
                    grade = Integer.parseInt(input.nextLine());
                } catch (NumberFormatException nfe) {   
                    System.out.println( " is not a valid grade.");
                    System.out.print("\n");
                    System.out.print("Enter grade (0-5): ");
                    grade = Integer.parseInt(input.nextLine());
                    }
            }                   
            System.out.print("OK");
        }

}

Answer 1

如果grade的值为9并且输入为“two”，这一行会发生什么？:

grade = Integer.parseInt(input.nextLine());

该行当然会抛出异常。但更重要的是，这一行不会将字符串值“two”分配给整数变量grade.

将输入存储在字符串中。像这样的东西:

String gradeInput = input.nextLine();
int grade = Integer.parseInt(gradeInput);

如果此操作抛出异常，则输入无效。如果它没有抛出异常，但 grade 超出了要求范围，则输入无效。但这两种情况都需要单独测试。

此外，请注意，您的第一个输入解析位于 try/catch 之外。所以它可能会完全失败。考虑这样的结构:

Scanner input = new Scanner(System.in);
String gradeInput;
int grade = -1;

while (grade < 0 || grade > 5) {
    try {
        System.out.print("Enter grade (0-5): ");
        gradeInput = input.nextLine();
        grade = Integer.parseInt(gradeInput);
        if (grade < 0 || grade > 5) {
            System.out.println(gradeInput + " is not a valid grade.");
            System.out.print("\n");
        }
    } catch (NumberFormatException nfe) {   
        System.out.println(gradeInput + " is not a valid grade.");
        System.out.print("\n");
    }
}

我想您可以通过多种方式将其组合在一起并进行重构，以尝试减少此处重复的几行。但重点是，您不想像当前代码那样重复接受输入并将其解析为整数的行。因为如果异常完全在 catch 中或在 try/catch 之外抛出，那么系统将失败。尝试从字符串中解析整数值需要位于try block 中。