检查字符串是否以 java 中的正则表达式以 2 个不同的数字结尾

Question

I must create a java program to check if the password is valid or not based on these conditions:

• to be at least 5 chars long or at most 12 chars long
• to start with an uppercase letter
• to end with two different digits
• to include at least one special character from the following: ! " # $ % & ' ( ) * + -
• to include at least one lowercase letter

This is what I have written so far, and I want to know what's the regular expression to check the second condition (that the password must end with 2 different digits)?

import java.util.Scanner;

public class PasswordValidation {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        
        System.out.println("Please enter a password");
        
        String password = sc.next();

        if (isValid(password)) {
   
            System.out.println("OK");
        } else {
  
            System.out.println("NO");
        }
        
    }

        public static boolean isValid (String password) {
            return password.matches ("^[A-Z](?=.*[a-z])(?=.*[!#$%&'()+-]).{5,12}$");
        }
        

}

Answer 1

Try using this regex pattern:

^(?=.*[a-z])(?=.*[!"#$%&'()*+-])[A-Z].{2,9}(\d)(?!\1)\d$

Java code:

String password = "Apple$123";
if (password.matches("(?=.*[a-z])(?=.*[!\"#$%&'()*+-])[A-Z].{2,9}(\\d)(?!\\1)\\d")) {
    System.out.println("MATCH");
}

This prints MATCH .

Here is an explanation of the regex pattern:

^                         from the start of the password
    (?=.*[a-z])           assert that a lowercase letter be present
    (?=.*[!"#$%&'()*+-])  assert that a special character be present
    [A-Z]                 password starts with uppercase letter
    .{2,9}                followed by any 2 to 9 characters (total 5 to 12)
    (\d)                  2nd to last character is any digit
    (?!\1)\d              last character is any digit other than previous one
$                         end of the password