Question

我需要一个java程序，询问0到2之间的数字。如果用户写0，程序结束。如果用户写1，则执行一个功能。如果用户写入2，则执行另一个功能。我还想用错误来处理错误＆＃34; java.lang.NumberFormatException＆＃34;，在这种情况下，再次询问用户一个数字，直到他写一个0到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;
            }
        }
    }

但是使用此代码时，catch（NumberFormatException z）不起作用，程序不会再次询问数字。

Answer 1

你从未真正抓住NumberFormatException。你的代码基本上是这样的：

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

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

您想要做的是：

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

    // etc
}

Answer 2

您可以将try / catch放在parseInt周围，如下所示：

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;
...

虽然 - 尝试 - 抓住Java

2 个答案: