正则表达式恰好匹配n次出现的字母和m次出现的数字

Question

I have to match an 8 character string, which can contain exactly 2 letters (1 uppercase and 1 lowercase), and exactly 6 digits, but they can be permutated arbitrarily.

So, basically:

• K82v6686 would pass
• 3w28E020 would pass
• 1276eQ900 would fail (too long)
• 98Y78k9k would fail (three letters)
• A09B2197 would fail (two capital letters)

I've tried using the positive lookahead to make sure that the string contains digits, uppercase and lowercase letters, but I have trouble with limiting it to a certain number of occurrences. I suppose I could go about it by including all possible combinations of where the letters and digits can occur:

(?=.*[0-9])(?=.*[A-Z])(?=.*[a-z]) ([A-Z][a-z][0-9]{6})|([A-Z][0-9][a-z][0-9]{5})| ... | ([0-9]{6}[a-z][A-Z])

But that's a very roundabout way of doing it, and I'm wondering if there's a better solution.

Answer 1

You can use

^(?=[^A-Z]*[A-Z][^A-Z]*$)(?=[^a-z]*[a-z][^a-z]*$)(?=(?:\D*\d){6}\D*$)[a-zA-Z0-9]{8}$

See the regex demo (a bit modified due to the multiline input). In Java, do not forget to use double backslashes (eg \\\\d to match a digit).

Here is a breakdown:

• ^ - start of string (assuming no multiline flag is to be used)
• (?=[^AZ]*[AZ][^AZ]*$) - check if there is only 1 uppercase letter (use \\p{Lu} to match any Unicode uppercase letter and \\P{Lu} to match any character other than that)
• (?=[^az]*[az][^az]*$) - similar check if there is only 1 lowercase letter (alternatively, use \\p{Ll} and \\P{Ll} to match Unicode letters)
• (?=(?:\\D*\\d){6}\\D*$) - check if there are six digits in a string (=from the beginning of the string, there can be 0 or more non-digit symbols ( \\D matches any character but a digit, you may also replace it with [^0-9] ), then followed by a digit ( \\d ) and then followed by 0 or more non-digit characters ( \\D* ) up to the end of string ( $ )) and then
• [a-zA-Z0-9]{8} - match exactly 8 alphanumeric characters.
• $ - end of string.

Following the logic, we can even reduce this to just

^(?=[^a-z]*[a-z][^a-z]*$)(?=(?:\D*\d){6}\D*$)[a-zA-Z0-9]{8}$

One condition can be removed as we only allow lower- and uppercase letters and digits with [a-zA-Z0-9] , and when we apply 2 conditions the 3rd one is automatically performed when matching the string (one character must be an uppercase in this case).

When using it with Java matches() method, there is no need to use ^ and $ anchors at the start and end of the pattern, but you still need it in the lookaheads:

String s = "K82v6686";
String rx = "(?=[^a-z]*[a-z][^a-z]*$)" +      // 1 lowercase letter check
            "(?=(?:\\D*\\d){6}\\D*$)" +       // 6 digits check
            "[a-zA-Z0-9]{8}";                 // matching 8 alphanum chars exactly
if (s.matches(rx)) {
    System.out.println("Valid"); 
} 

Answer 2

Pattern.matches(".*[A-Z].*", s) &&
Pattern.matches(".*[a-z].*", s) &&
Pattern.matches(".*(\\D*\\d){6}.*", s) &&
Pattern.matches(".{8}", s)

As we need an alternating automaton to be created for this task, it's much simpler to use a conjunction of regexps for constituent types of character.

We require it to have at least one lowercase letter, one uppercase letter and 6 digits, which three classes are mutually exclusive. And with the last condition we require the length of string to be exactly the sum of these numbers in such a way leaving no room for extra characters beyond the desired types. Of course we may say s.lenght() == 8 as the last condition term but this would break the style :).

Answer 3

词法排序字符串，然后匹配^(?:[az][AZ]|[AZ][az])[0-9]{6}$ 。