超过“9999”时将数字字符串值重置为“1”

Question

如何将字符串编号 9999 重置为 0001，限制为 4 位：

String currentNumber = "9999";

String incremented = String.format("%0" + currentNumber.length() + "d",

                Integer.parseInt(currentNumber) + 1);

上面的结果会给我 10000。

Answer 1

使用模运算符：String currentNumber = "9999";String incremented = String.format("%0" + currentNumber.length() + "d",                (Integer.parseInt(currentNumber) + 1) % 9999);这就是当您使用正确的打字机时的样子currentNumber：Integer currentNumber = 9999;Integer moduledNumber = (currentNumber + 1) % 9999;String incremented = String.format("%0" + currentNumber.toString().length() + "d", moduledNumber);

Answer 2

您可以为此创建自定义方法String increment(String num, int incrementBy) //incrementBy  is an integer by which you want to increment{      int numInt=Integer.parseInt(num);    numInt+=incrementBy;    if (numInt>9999) {        return String.valueOf(numInt%9999);    }    return String.valueOf(numInt);}

Answer 3

为什么不使用 if 语句？尝试这个if (Integer.parseInt(currentNumber) > Integer.parseInt(maxNumber))currentNumber = "0001"希望它会帮助你