Question

我希望用户输入一个由以下代码扫描的号码：

scanner.nextInt();

如果用户输入字符串，程序会抛出InputMismatchException，这很明显。我想以这样的方式捕获异常，即程序提示用户输入输入，直到用户输入整数值。

Scanner scanner = new Scanner(System.in);
while(true) {
    try {
        System.out.println("Please enter a number: ");
        int input = scanner.nextInt();
        System.out.println(input);
        //statements
        break;
    }
    catch(InputMismatchException | NumberFormatException ex ) {
        continue;
    }
}

如果输入字符串，此代码会创建一个无限循环。

Answer 1

我的问题的答案如下：

Scanner scanner = new Scanner(System.in);
while(true) {
    try {
        System.out.println("Please enter a number: ");
        int input = scanner.nextInt();
        System.out.println(input);
        //statements
        break;
    }
    catch(InputMismatchException | NumberFormatException ex ) {
        scanner.next();//new piece of code which parses the wrong input and clears the //scanner for new input
        continue;
    }
}

Answer 2

将Scanner scanner = new Scanner(System.in);放在while循环中。

Scanner scanner;
while(true) {    
    try {
        System.out.println("Please enter a number: ");
        scanner = new Scanner(System.in);
        int input = scanner.nextInt();
        System.out.println(input);
        //statements
        break;
    }
    catch(InputMismatchException | NumberFormatException ex ) {
        System.out.println("I said a number...");
    }
}

Answer 3

这个怎么样？

while(true) {    
    try {
        System.out.println("Please enter a number: ");
        Scanner scanner = new Scanner(System.in);
        int input = scanner.nextInt();
        System.out.println("\n\nEntered number is : " + input);
        break;
    } catch(InputMismatchException | NumberFormatException ex ) {
        System.out.println("\n\nInput was not a number. Please enter number again : ");
    } catch(Exception e ) {
        System.out.println("\n\nException caught :: " + e);
    }
}

我还删除了continue语法，因为不需要这些语法。

捕获异常后请求输入

3 个答案: