Regex Lookahead and lookbehind no more than one digit

Question

I am looking to create a RegEx template

• 8 characters [a-zA_Z]
• must contain only one digit anywhere on the line

I created this template:

 ^(?=.*[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

Instead, you can use:

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

The first part will claim that the match contains 8 alphabets or numbers. Once this is ensured, the second part ensures that there is only one number in the match.

EDIT: Explanation:

• Anchors ^ and $ denote the beginning and end of a line.
• (?=[0-9a-zA-Z]{8}) states that the match contains 8 alphabets or numbers.
• [^\d]*\d[^\d]* will mean that there is only one digital character and the remaining non-character characters. Since we have already claimed that the input contains numbers or alphabets, characters without numbers here are alphabets.

Answer 2

If you want a non-regular expression, 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; }