Question

我试图以两个不同的整数形式输入输入，这些整数用空格ex 14 2分隔。

如果输入是double而不是int，一个字符，或者基本上不是由空格分隔的两个整数，我找不到提示用户再次尝试的方法。

我真的不想使用exceptions and try/catch语句，因为我不理解它们。

我宁愿接受字符串[] arr并使用split（“”）或只做console.nextInt();。任何反馈都会很棒。我知道它不漂亮。

Scanner console = new Scanner(System.in);
boolean ok = true;
boolean ok2 = true;
while (ok = true){
for(int i = 0; i<=2; i++){

if(!console.hasNextInt()){
    kb.next();
    System.out.println("Bad input, try again");
    ok2 false;
}
else if(i = 0 && ok2 = true){
    int a = console.hasNextInt();
}
else if(i = 1 && ok2 = true){
    int b = console.hasNextInt();
    ok = false; //means there are two #s and the program can continue with a and b   

}

else{
}
 }

Answer 1

以下代码接受输入，直到只获得正数。否定与零数字由while循环条件处理。在另一个while循环条件下使用!sc.hasNextInt()处理字符。

       Scanner sc = new Scanner(System.in);
        int number;
        do {
            System.out.println("Please enter a positive number : ");
            while (!sc.hasNextInt()) {
                System.out.println("That is not a valid number.");
                sc.next(); 
            }
            number = sc.nextInt();
        } while (number <= 0);
        System.out.println("Recieved a Positive number = " + number+". Thanks!");

输出 -

Please enter a positive number : 
-1
Please enter a positive number : 
0
Please enter a positive number : 
n
That is not a valid number.
6
Recieved a Positive number = 6. Thanks!

Answer 2

我正在努力编写尽可能简单易用的代码，

Scanner console = new Scanner(System.in);
System.out.print("Enter two numbers  ");

String str = console.readLine();

String values[] = str.split(" ");

int n1, n2;

try{
    n1 = Integer.parseInt(values[0]);
    //Convert first String value to int, if it will be anything else than int
    // then it will throw the NumberFormatException
}catch(NumberFormatException e){
    System.out.println("First number is not integer");
}

try{
    n2 = Integer.parseInt(values[1]);
    //Convert second String value to int, if it will be anything else than int
    // then it will throw the NumberFormatException
}catch(NumberFormatException e){
    System.out.println("Second number is not integer");
}

注意

此代码基于以下假设：仅用户输入的两个元素。不超过两个。在这种情况下，需要更改代码。

Answer 3

使用Scanner＃hasNextInt（）验证输入的另一种解决方案：

    final Scanner console = new Scanner(System.in);
    final List<Integer> input = new ArrayList<Integer>();

    while (true)
    {
        System.out.print("Please enter an integer : ");

        if (console.hasNextInt())
        {
            input.add(console.nextInt());
        }
        else
        {
            System.out.println("Bad input, try again ...");
            console.next();
        }
        if (input.size() >= 2)
        {
            break;
        }
    }

    System.out.println("\n");
    System.out.println("First integer entered was : " + input.get(0));
    System.out.println("Second integer entered was : " + input.get(1));

如何管理不正确的输入

3 个答案: