Regex Lookahead and lookbehind at most one digit

Question

I'm looking for create RegEx pattern

• 8 characters [a-zA_Z]
• must contains only one digit in any place of string

I created this pattern:

^(?=.*[0-9].*[0-9])[0-9a-zA-Z]{8}$

This pattern works fine but i want only one digit allowed. Example:

aaaaaaa6   match
aaa7aaaa   match

aaa88aaa   don't match
aaa884aa   don't match
aaawwaaa   don't match

Answer 1

You could instead use:

^(?=[0-9a-zA-Z]{8})[^\d]*\d[^\d]*$

The first part would assert that the match contains 8 alphabets or digits. Once this is ensured, the second part ensures that there is only one digit in the match.

EDIT: Explanation:

• The anchors ^ and $ denote the start and end of string.
• (?=[0-9a-zA-Z]{8}) asserts that the match contains 8 alphabets or digits.
• [^\d]*\d[^\d]* would imply that there is only one digit character and remaining non-digit characters. Since we had already asserted that the input contains digits or alphabets, the non-digit characters here are alphabets.

Answer 2

If you want a non regex solution, I wrote this for a small project :

public static bool ContainsOneDigit(string s)
{
    if (String.IsNullOrWhiteSpace(s) || s.Length != 8)
        return false;
    int nb = 0;
    foreach (char c in s)
    {
        if (!Char.IsLetterOrDigit(c))
            return false;
        if (c >= '0' && c <= '9') // just thought, I could use Char.IsDigit() here ...
            nb++;
    }
    return nb == 1;
}