Question

我正在尝试创建一个代码，其中应该输入一个int，然后如果int不在9到99之间则有异常，如果输入double而不是int则是另一个异常，然后是第三个异常输入字符串。我该怎么做呢？我有到目前为止我所拥有的，但我不知道如何纠正它。感谢

public static void main(String[] args) {
    Scanner input = new Scanner(System.in);
    boolean correct = true;
    do {
        try {
            System.out.println("Enter an Integer between 9 and 99");
            int number = input.nextInt();
            if (number >= 9 && number <= 99) {
                System.out.println("Thank you, Initialization completed");
                correct = false;
            } else if (number < 9 || number > 99) {
                throw new Exception("Integer is not within the range");
            }
            if (input.hasNextDouble()) {
                throw new Exception("Integer not entered");
            } else {
                correct = false;
            }
            if (input.hasNext("")) {
                throw new NumberFormatException("Integer not entered");
            } else {
                correct = false;
            }
        } // check for range
        catch (Exception e1) {
            System.out.println("Number is not within 9 and 99");
            System.out.println();
            input.nextLine();
        } catch (Exception e2) {
            System.out.println("An integer was not entered");
            System.out.println();
            input.nextLine();
        } catch (NumberFormatException e3) {
            System.out.println("An integer was not entered");
            System.out.println();
            input.nextLine();
        }
    } while (correct);
}

Answer 1

方法 .getMessage（）返回构造函数中给出的字符串：

throw new Exception("HERE");

当您捕获Exception时，您还会捕获NumberFormatException，InputMismatchException等。所以你最后必须抓住更广泛的。

catch (NumberFormatException e3) { // Precisier goes first
    System.out.println("An integer was not entered");
    System.out.println();
    input.nextLine();
}
catch (Exception e1) {
    System.out.println(e1.getMessage());
    System.out.println();
    input.nextLine();
}

int的三个例外

1 个答案: