虽然-尝试-赶上Java

Question

我需要一个Java程序，要求输入0到2之间的数字。如果用户输入0，则该程序结束。 如果用户写1，它将执行一个功能。 如果用户写2，它将执行另一个功能。 我还想用一条消息处理错误“ java.lang.NumberFormatException”，在这种情况下，请再次询问用户一个数字，直到他写一个介于0和2之间的数字。

我用

public static void main(String[] args) throws IOException {
    int number = 0;
    boolean numberCorrect = false;
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));



        while (numberCorrect == false){
            System.out.println("Choose a number between 0 and 2");
            String option = br.readLine();
            number = Integer.parseInt(option);

            try {
                switch(option) {
                case "0":
                    System.out.println("Program ends");
                    numberCorrect = true;
                    break;
                case "1":
                    System.out.println("You choose "+option);
                    functionA();
                    numberCorrect = true;
                    break;
                case "2":
                    System.out.println("You choose  "+option);
                    functionB();
                    numberCorrect = true;
                    break;
                default:
                    System.out.println("Incorrect option");
                    System.out.println("Try with a correct number");
                    numberCorrect = false;
                }   
            }catch(NumberFormatException z) {
                System.out.println("Try with a correct number");
                numberCorrect = false;
            }
        }
    }

但是使用此代码，catch（NumberFormatException z）不起作用，并且程序不再要求输入数字。

Answer 1

您从不真正在这里捕获NumberFormatException 。 您的代码基本上可以做到：

while (...) {
    // this can throw NumberFormatException
    Integer.parseInt(...)

    try {
        // the code in here cannot
    } catch (NumberFormatException e) {
        // therefore this is never reached
    }
}

您要在这里做的是：

while (!numberCorrect) {
    line = br.readLine();
    try {
        number = Integer.parseInt(line);
    } catch (NumberFormatException ignored) {
        continue;
    }

    // etc
}

Answer 2

您可以像这样将try / catch放在parseInt周围：

while (numberCorrect == false){
   System.out.println("Choose a number between 0 and 2");
   String option = br.readLine();

   try {
        number = Integer.parseInt(option);
    }catch(NumberFormatException z) {
       System.out.println("Try with a correct number");
       numberCorrect = false;
       option = "-1";
    }

    switch(option) {
        case "0":
        System.out.println("Program ends");
        numberCorrect = true;
        break;
...