正则表达式验证密码强度

9 回答

• 0

  对于PHP，这很好用！

  if(preg_match("/^(?=(?:[^A-Z]*[A-Z]){2})(?=(?:[^0-9]*[0-9]){2}).{8,}$/", 
   'CaSu4Li8')){
      return true;
   }else{
      return fasle;
   }
  

  在这种情况下，结果是真的

  @ridgerunner的问题

  回复于 2024-04-23T15:34:30+08:00
• 367

  您可以使用正向预测断言进行这些检查：

  ^(?=.*[A-Z].*[A-Z])(?=.*[!@#$&*])(?=.*[0-9].*[0-9])(?=.*[a-z].*[a-z].*[a-z]).{8}$
  

  Rubular link

  说明：

  ^                         Start anchor
  (?=.*[A-Z].*[A-Z])        Ensure string has two uppercase letters.
  (?=.*[!@#$&*])            Ensure string has one special case letter.
  (?=.*[0-9].*[0-9])        Ensure string has two digits.
  (?=.*[a-z].*[a-z].*[a-z]) Ensure string has three lowercase letters.
  .{8}                      Ensure string is of length 8.
  $                         End anchor.
  

  回复于 2024-04-23T15:34:30+08:00
• 2

  您可以使用零长度正向预测来分别指定每个约束：

  (?=.{8,})(?=.*\p{Lu}.*\p{Lu})(?=.*[!@#$&*])(?=.*[0-9])(?=.*\p{Ll}.*\p{Ll})
  

  如果您的正则表达式引擎不支持 \p 表示法并且纯ASCII足够，那么您可以将 \p{Lu} 替换为 [A-Z] 并将 \p{Ll} 替换为 [a-z] .

  回复于 2024-04-23T15:34:30+08:00
• 9

  上面给出的答案是完美的，但我建议使用多个较小的正则表达式而不是大的正则表达式 .
  拆分长正则表达式有一些优点：

  • 易于书写和阅读

  • 易于调试

  • 容易添加/删除部分正则表达式

  通常这种方法保留代码 easily maintainable .

  话虽如此，我分享了一段我在 Swift 中编写的代码：

  struct RegExp {
  
      /**
       Check password complexity
  
       - parameter password:         password to test
       - parameter length:           password min length
       - parameter patternsToEscape: patterns that password must not contains
       - parameter caseSensitivty:   specify if password must conforms case sensitivity or not
       - parameter numericDigits:    specify if password must conforms contains numeric digits or not
  
       - returns: boolean that describes if password is valid or not
       */
      static func checkPasswordComplexity(password password: String, length: Int, patternsToEscape: [String], caseSensitivty: Bool, numericDigits: Bool) -> Bool {
          if (password.length < length) {
              return false
          }
          if caseSensitivty {
              let hasUpperCase = RegExp.matchesForRegexInText("[A-Z]", text: password).count > 0
              if !hasUpperCase {
                  return false
              }
              let hasLowerCase = RegExp.matchesForRegexInText("[a-z]", text: password).count > 0
              if !hasLowerCase {
                  return false
              }
          }
          if numericDigits {
              let hasNumbers = RegExp.matchesForRegexInText("\\d", text: password).count > 0
              if !hasNumbers {
                  return false
              }
          }
          if patternsToEscape.count > 0 {
              let passwordLowerCase = password.lowercaseString
              for pattern in patternsToEscape {
                  let hasMatchesWithPattern = RegExp.matchesForRegexInText(pattern, text: passwordLowerCase).count > 0
                  if hasMatchesWithPattern {
                      return false
                  }
              }
          }
          return true
      }
  
      static func matchesForRegexInText(regex: String, text: String) -> [String] {
          do {
              let regex = try NSRegularExpression(pattern: regex, options: [])
              let nsString = text as NSString
              let results = regex.matchesInString(text,
                  options: [], range: NSMakeRange(0, nsString.length))
              return results.map { nsString.substringWithRange($0.range)}
          } catch let error as NSError {
              print("invalid regex: \(error.localizedDescription)")
              return []
          }
      }
  }
  

  回复于 2024-04-23T15:34:30+08:00
• 1

  我建议补充一下

  (?!.*pass|.*word|.*1234|.*qwer|.*asdf) exclude common passwords
  

  回复于 2024-04-23T15:34:30+08:00
• 0

  codaddict的解决方案工作正常，但这个效率更高一些:( Python语法）

  password = re.compile(r"""(?#!py password Rev:20160831_2100)
      # Validate password: 2 upper, 1 special, 2 digit, 1 lower, 8 chars.
      ^                        # Anchor to start of string.
      (?=(?:[^A-Z]*[A-Z]){2})  # At least two uppercase.
      (?=[^!@#$&*]*[!@#$&*])   # At least one "special".
      (?=(?:[^0-9]*[0-9]){2})  # At least two digit.
      .{8,}                    # Password length is 8 or more.
      $                        # Anchor to end of string.
      """, re.VERBOSE)
  

  被否定的字符类在一个步骤中消耗所有字符，需要零回溯 . （点星解决方案工作得很好，但确实需要一些回溯 . ）当然，对于密码等短目标字符串，这种效率提升可以忽略不计 .

  回复于 2024-04-23T15:34:30+08:00
• 0

  import re
  
  RegexLength=re.compile(r'^\S{8,}$')
  RegexDigit=re.compile(r'\d')
  RegexLower=re.compile(r'[a-z]')
  RegexUpper=re.compile(r'[A-Z]')
  
  
  def IsStrongPW(password):
      if RegexLength.search(password) == None or RegexDigit.search(password) == None or RegexUpper.search(password) == None or RegexLower.search(password) == None:
          return False
      else:
          return True
  
  while True:
      userpw=input("please input your passord to check: \n")
      if userpw == "exit":
          break
      else:
          print(IsStrongPW(userpw))
  

  回复于 2024-04-23T15:34:30+08:00
• 0

  另一种方案：

  import re
  
  passwordRegex = re.compile(r'''(
      ^(?=.*[A-Z].*[A-Z])                # at least two capital letters
      (?=.*[!@#$&*])                     # at least one of these special c-er
      (?=.*[0-9].*[0-9])                 # at least two numeric digits
      (?=.*[a-z].*[a-z].*[a-z])          # at least three lower case letters
      .{8,}                              # at least 8 total digits
      $
      )''', re.VERBOSE)
  
  def userInputPasswordCheck():
      print('Enter a potential password:')
      while True:
          m = input()
          mo = passwordRegex.search(m) 
          if (not mo):
             print('''
  Your password should have at least one special charachter,
  two digits, two uppercase and three lowercase charachter. Length: 8+ ch-ers.
  
  Enter another password:''')          
          else:
             print('Password is strong')
             return
  userInputPasswordCheck()
  

  回复于 2024-04-23T15:34:30+08:00
• 6

  密码必须满足以下4个复杂性规则中的至少3个，

  [至少1个大写字母（A-Z）至少1个小写字符（a-z）至少1个数字（0-9）至少1个特殊字符 - 不要忘记将空格视为特殊字符]

  至少10个字符

  最多128个字符

  连续不超过2个相同的字符（例如，111不允许）

  '^（？！ . （ . ）\ 1 {2}）（（？= . [az]）（？= . [AZ]）（？= . [0-9]）|（？= . [az] ）（= [AZ]）（= [^A-ZA-Z0-9]）|？？？？？（= [AZ]）（= [0-9]）（= [^一个-Za-Z0-9]）|？（= [AZ]）（= [0-9]）（= * [^A-ZA-Z0-9]））{10127} $”？？ .

  （？！ . *（ . ）\ 1 {2}）

  （？= [A-Z]）（？= [A-Z]）（？= . * [0-9]）

  （？= [A-Z]）（？= [A-Z]）（？= . * [^A-ZA-Z0-9]）

  （？= [A-Z]）（？= . [0-9]）（？= . * [^A-ZA-Z0-9]）

  （？= [A-Z]）（？= . [0-9]）（？= . * [^A-ZA-Z0-9]）

  {} 10.127

  回复于 2024-04-23T15:34:30+08:00