正则表达式查找最后 4 位数字和 1 个特殊字符

Question

How do I find the last 4 digits and 1 @ using regx?

@ character will never be the last character.

123@1234@1234 567@1 ---> returns 567@1

123@1234@1234 56@12 ---> returns 56@12

123@1234@1234 5@123 ---> returns 5@123

123@1234@1234@ 1234 ---> returns 1234

Thanks for the comments bringing up these variations

Update:

123@123 4@23@1 —> returns 4@23@1

12@123@1 2@1@1@1 —> returns 2@1@1@1

Answer 1

This regex seems working with your test cases:

(\d{4}|(?:\d@*){4})$

Demo & explanation

Answer 2

You can simply loop througt the string starting from the end, and count the digit. Once you reach the number of digit wanted, substring from its index.

private static string LastNDigit(string input, int nbDigit)
{
    var digitCount = 0;
    var sIndex = 0;           

    for (int i = input.Length - 1; i >= 0; i--)
    {
        if (char.IsDigit(input[i])) digitCount++;
        if (digitCount == nbDigit) {
            sIndex = i;
            break;
        }
    }

    return input.Substring(sIndex);
}

Usage:

var inputs = new[]{
        "123@1234@1234567@1",
        "123@1234@123456@12",
        "123@1234@12345@123",
        "123@1234@1234@1234"
};

foreach (var input in inputs)
{
    Console.WriteLine(input + " => " + LastNDigit(input, 4));
}

live Demo