Java扫描器前瞻

Question

有没有办法实现扫描仪。hasNextInt（）以便它不仅可以检查i+1令牌，还可以检查i+2令牌？我正在寻找一种方法来检查两个输入是否都是数字，而不必打印两次相同的错误消息

输入应为以下格式：

1 2

我试图提醒用户，如果输入的是表单

1 a, a 1, a a, aa aa, 1a 1, etc...

我希望它是什么样子：（不这样工作）

            int pile1, pile2;

        // the second call to reader.hasNextInt() would be verifying pile2 to be int
        if (reader.hasNextInt() && reader.hasNextInt())
        {
            pile1 = reader.nextInt();
            pile2 = reader.nextInt();
        }
        else
        {
            System.out.println("Your input is malformed. Try again");
        }

我目前拥有的：

                int pile1, pile2;
                if (reader.hasNextInt())
                    pile1 = reader.nextInt();
                else
                {
                   System.out.println("Your input is malformed. Try again");
                    return;
                }

                if (reader.hasNextInt())
                    pile2 = reader.nextInt();
                else
                {
                   System.out.println("Your input is malformed. Try again");
                    return;
                }

Answer 1

你可以用

hasNext(String pattern) this method returns true if the next token matches the pattern constructed from the specified string.

public static void main(String[] args) {

    Scanner reader = new Scanner(System.in);
    List<Integer> data = new ArrayList<Integer>();
    System.out.println("input count of numbers to inputed:\t");
    int limit = 0;
    if (reader.hasNext("\\d+")) {
        reader.nextInt();
    } else {
        System.out.println("wrong input");
        return;
    }
    System.out.println("please input data:\t");

    for (int i = 0; i < limit; i++) {

        if (reader.hasNext("\\d+")) {
            data.add(reader.nextInt());
        } else {
            System.out.println("Your input is malformed. Try again");
            break;
        }
    }

    reader.close();

}

假设：由于您需要停止扫描，所以必须设置预定义的限制或循环的特殊退出条件