首页 > 解决方案 > 在java中使用正则表达式检查字符串是否以2个不同的数字结尾

问题描述

我必须创建一个 java 程序来根据这些条件检查密码是否有效:

这是我到目前为止所写的,我想知道检查第二个条件的正则表达式是什么(密码必须以 2 个不同的数字结尾)?

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])(?=.*[!#$%&'()+-]).{5,12}$");
        }
        

}

标签: java

解决方案


尝试使用此正则表达式模式:

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

Java代码:

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

这打印MATCH

以下是正则表达式模式的解释:

^                         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

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