2 个字母和 6 个数字的正则表达式

Question

I have a password rule to be follow, required the password must contain 2 alphabet (upper or lowercase) & 6 numbers, and the whole length is expected to equal 8 symbols.

Here's the samples that should pass:

a123456b
1a2b3456
aa123456
123ab456

And the samples that should fail:

1abcdefg2
a1234567b
abcdefgh
12345678

And I need a regular expression to fit this rule.

Answer 1

Edit : TL;DR: Regex are not the tool for the job, and the job doesn't need to be done.

Regex is not the solution to the problem, and AFAIK it is impossible to use it to solve your problem. (But don't take my word for it, I bet someone wrote some funky extension to an esoteric Regex engine that does just this)

If it is possible, the solution will be unreadable. It is better to just check the result of Count .

Something along the lines of (haven't actually compiled this):

bool IsPasswordValid(string pw)
{
    return pw.Length == 8 && 
            pw.Count(char.IsNumber) == 6 &&
            pw.Count(char.IsLetter) == 2;
}

Notes :

Regular Expressions are (is?) a tool that is used to check a string against a pattern. The simpler the pattern, the easier it is to write a regex for it. Since you want your password to have 2-chars of a set anywhere in the string, and 6 character of another set also anywhere in the string, it is very hard (or even impossible) to create such a pattern.

If your problem was "I want the password to start with 2 digits and end with 6 letters", the answer would've been trivial, but since the digits and letters can be separate and anywhere in the string, it is not as trivial.

Also, most of the time, enforcing password patterns does not increase security. If you have to do anything, enforce a sane minimum length ans stop there.

Answer 2

A regex approach can be multiple, here are 2: 1) require the legnth of 8 chars with a lookahead like (?=.{8}$) and use a consuming pattern allowing zero or more digits before a letter twice, and then match zero or more letters, 2) match 8 letters or digits, but use a lookahead restriction to match 2 alphabets.

Example 1:

^(?=.{8}$)(?:\d*[a-z]){2}\d*$

Details :

• ^ - start of string
• (?=.{8}$) - the length should be 8 non-newline symbols only
• (?:\\d*[az]){2} - 2 sequences of 0+ digits followed with 1 letter
• \\d* - 0+ digits
• $ - end of string.

See the regex demo

Example 2:

^(?=(?:\d*[a-z]){2}\d*$)[\da-z]{8}$

Details :

• ^ - start of string
• (?=(?:\\d*[az]){2}\\d*$) - there must be 2 sequences of 0+ digits and a letter followed with 0+ digits up to the end of string
• [\\da-z]{8} - 8 digits or letters
• $ - end of string.

See the regex demo 2

Note that to enable case insensitive mode, you can prepend the pattern with (?i) / RegexOptions.IgnoreCase flag, and if you need to only match ASCII digits, use RegexOptions.ECMAScript or replace \\d with [0-9] .

Answer 3

Actually,there is a nice way to do it:

    public static bool CheckPasswordRule(string password)
    {
        var isRuleAdhered = (Regex.Matches(password, "\\d").Count == 6 && Regex.Matches(password, "[a-z]").Count == 2);

        return isRuleAdhered;
    }

If you still want to not have 6 digits one after another you can modify the code to:

    public static bool CheckPasswordRule(string password)
    {
        var isRuleAdhered= false;
        isRuleAdhered = Regex.Matches(password, "\\d").Count == 6 && Regex.Matches(password, "[a-z]").Count == 2 && !Regex.IsMatch(password, "\\d{6}");
        return isRuleAdhered;
    }