java - while true try catch 嵌套

Question

我是 Java 新手。我希望代码在用户输入的类型错误的地方重复，而不是从头开始。当“输入 b:”或“输入 c:”时，它会返回到开头“输入 a:”。我希望它仅重复，其中用户输入是 a、b、c。提前致谢。

public static void main(String[] args)
{
Scanner sc = new Scanner(System.in);
boolean itsANumber = true;
while (itsANumber)
{
    System.out.print("Enter a: ");

    try
    {
    a = Double.parseDouble(sc.nextLine());

    System.out.print("Enter b: ");
    try
    {
        b = Double.parseDouble(sc.nextLine());

        System.out.print("Enter c: ");
        try
        {
        c = Double.parseDouble(sc.nextLine());
        if (a == 0)
        {
            aZero();
        } else

        {
            aNotZero();
        }

        } catch (NumberFormatException nfe) 
        {
        System.out
            .println("That's not a number, please try again!");
        }

    } catch (NumberFormatException nfe) 
    {
        System.out
            .println("That's not a number, please try again!");
    }

    } catch (NumberFormatException nfe) 
    {
    System.out.println("That's not a number, please try again!");
    }
}

}

Answer 1

引入一个请求号码的方法并调用它三次。在该方法内部，您将拥有带有 try-catch 的 while 循环。

public static void main(String... args) {
   Scanner sc = new Scanner(System.in);
   double
     a = askForDouble(sc, "a"),
     b = askForDouble(sc, "b"),
     c = askForDouble(sc, "c");
}
static double askForDouble(Scanner sc, String varName) {
   for (;/*ever*/;) {
     System.out.format("Enter %s: ", varName);
     System.out.flush();
     try {
       return Double.parseDouble(sc.nextLine());
     } catch (NumberFormatExcetpion() {
       System.out.println("That's not a number, please try again!");
     }
   }
}