Regular expression to check if password is "8 characters including 1 uppercase letter, 1 special character, alphanumeric characters"

( # Start of group (?=.*\d) # must contain at least one digit (?=.*[A-Z]) # must contain at least one uppercase character (?=.*\W) # must contain at least one special symbol . # match anything with previous condition checking {8,8} # length is exactly 8 characters ) # End of group 

In one line:

((?=.*\d)(?=.*[A-Z])(?=.*\W).{8,8}) 

Edit 2019-05-28:

You need to match entire input string. So, you can enclose the regex between ^ and $ to prevent accidentally assuming partial matches as matching entire input:

^((?=.*\d)(?=.*[A-Z])(?=.*\W).{8,8})$ 

Sources:

• Password matching expression

• Password Strength Validation with Regular Expressions

So many answers.... all bad!

Regular expressions don't have an AND operator, so it's pretty hard to write a regex that matches valid passwords, when validity is defined by something AND something else AND something else...

But, regular expressions do have an OR operator, so just apply DeMorgan's theorem, and write a regex that matches invalid passwords.

anything with less than 8 characters OR anything with no numbers OR anything with no uppercase OR anything with no special characters

So:

^(.{0,7}|[^0-9]*|[^A-Z]*|[a-zA-Z0-9]*)$ 

If anything matches that, then it's an invalid password.

The answer is to not use a regular expression. This is sets and counting.

Regular expressions are about order.

In your life as a programmer you will asked to do many things that do not make sense. Learn to dig a level deeper. Learn when the question is wrong.

The question (if it mentioned regular expressions) is wrong.

Pseudocode (been switching between too many languages, of late):

if s.length < 8: return False nUpper = nLower = nAlphanum = nSpecial = 0 for c in s: if isUpper(c): nUpper++ if isLower(c): nLower++ if isAlphanumeric(c): nAlphanum++ if isSpecial(c): nSpecial++ return (0 < nUpper) and (0 < nAlphanum) and (0 < nSpecial) 

Bet you read and understood the above code almost instantly. Bet you took much longer with the regex, and are less certain it is correct. Extending the regex is risky. Extended the immediate above, much less so.

Note also the question is imprecisely phrased. Is the character set ASCII or Unicode, or ?? My guess from reading the question is that at least one lowercase character is assumed. So I think the assumed last rule should be:

return (0 < nUpper) and (0 < nLower) and (0 < nAlphanum) and (0 < nSpecial) 

(Changing hats to security-focused, this is a really annoying/not useful rule.)

Learning to know when the question is wrong is massively more important than clever answers. A clever answer to the wrong question is almost always wrong.

As an example how this could be done with a readable/maintainable regex.

For a longer regex you should always use RegexOptions.IgnorePatternWhitespace to allow whitespace and comments in the expression for better readability.

String[] passwords = { "foobar", "Foobar", "Foobar1", "Fooobar12" }; foreach (String s in passwords) { Match password = Regex.Match(s, @" ^ # Match the start of the string (?=.*\p{Lu}) # Positive lookahead assertion, is true when there is an uppercase letter (?=.*\P{L}) # Positive lookahead assertion, is true when there is a non-letter \S{8,} # At least 8 non whitespace characters $ # Match the end of the string ", RegexOptions.IgnorePatternWhitespace); if (password.Success) { Console.WriteLine(s + ": valid"); } else { Console.WriteLine(s + ": invalid"); } } Console.ReadLine(); 

If you need only one upper case and special character then this should work:

@"^(?=.{8,}$)(?=[^A-Z]*[A-Z][^A-Z]*$)\w*\W\w*$" 

The regular expression you was looking for is: /^(?=.*[a-z])(?=.*[A-Z])(?=.*[0-9])(?=.*[!@#\$%\^&\*\[\]"\';:_\-<>\., =\+\/\\]).{8,}$/u.

Example and test: http://regexr.com/3fhr4

This question starts to be viral and a lot of interesting suggestion appeared.

Yes, writing by hand is difficult. So an easier solution is to use a template. Although the resulted regex may not be most optimal, it will be easier to maintain and/or change, and the user will have a better control over the result. It is possible that I missed something, so any constructive criticism will be helpful.

This links might be interesting: match at least 2 digits 2 letters in any order in a string, Regular Expression Language, Capturing groups

I'm using this template (?=(?:.*?({type})){({count})}) based on all of the regex I saw in SO. The next step is replacing the needed pattern ( number, special character ... ) and adding configuration for length.

I've made a little class for composing the regex PasswordRegexGenerator.cs An example:

string result = new PasswordRegexGenerator ( ) .UpperCase ( 3, -1 ) // ... {3,} .Number ( 2, 4 ) // ... {2,4} .SpecialCharacter ( 2 ) // ... {2} .Total ( 8,-1 ) .Compose ( ); /// <summary> /// Generator for regular expression, validating password requirements. /// </summary> public class PasswordRegexGenerator { private string _elementTemplate = "(?=(?:.*?({type})){({count})})"; private Dictionary<string, string> _elements = new Dictionary<string, string> { { "uppercase", "[A-Z]" }, { "lowercase", "[a-z]" }, { "number", @"\d" }, { "special", @"\W" }, { "alphanumeric", @"\w" } }; private StringBuilder _sb = new StringBuilder ( ); private string Construct ( string what, int min, int max ) { StringBuilder sb = new StringBuilder ( _elementTemplate ); string count = min.ToString ( ); if ( max == -1 ) { count += ","; } else if ( max > 0 ) { count += "," + max.ToString(); } return sb .Replace ( "({type})", what ) .Replace ( "({count})", count ) .ToString ( ); } /// <summary> /// Change the template for the generation of the regex parts /// </summary> /// <param name="newTemplate">the new template</param> /// <returns></returns> public PasswordRegexGenerator ChangeRegexTemplate ( string newTemplate ) { _elementTemplate = newTemplate; return this; } /// <summary> /// Change or update the regex for a certain type ( number, uppercase ... ) /// </summary> /// <param name="name">type of the regex</param> /// <param name="regex">new value for the regex</param> /// <returns></returns> public PasswordRegexGenerator ChangeRegexElements ( string name, string regex ) { if ( _elements.ContainsKey ( name ) ) { _elements[ name ] = regex; } else { _elements.Add ( name, regex ); } return this; } #region construction methods /// <summary> /// Adding number requirement /// </summary> /// <param name="min"></param> /// <param name="max"></param> /// <returns></returns> public PasswordRegexGenerator Number ( int min = 1, int max = 0 ) { _sb.Append ( Construct ( _elements[ "number" ], min, max ) ); return this; } public PasswordRegexGenerator UpperCase ( int min = 1, int max = 0 ) { _sb.Append ( Construct ( _elements[ "uppercase" ], min, max ) ); return this; } public PasswordRegexGenerator LowerCase ( int min = 1, int max = 0 ) { _sb.Append ( Construct ( _elements[ "lowercase" ], min, max ) ); return this; } public PasswordRegexGenerator SpecialCharacter ( int min = 1, int max = 0 ) { _sb.Append ( Construct ( _elements[ "special" ], min, max ) ); return this; } public PasswordRegexGenerator Total ( int min, int max = 0 ) { string count = min.ToString ( ) + ( ( max == 0 ) ? "" : "," + max.ToString ( ) ); _sb.Append ( ".{" + count + "}" ); return this; } #endregion public string Compose () { return "(" + _sb.ToString ( ) + ")"; } } 

You can use below class for validation:

public class PasswordValidator{ private Pattern pattern; private Matcher matcher; private static final String PASSWORD_PATTERN = "((?=.*\\d)(?=.*[a-z])(?=.*[A-Z])(?=.*[@#$%]).{6,20})"; public PasswordValidator(){ pattern = Pattern.compile(PASSWORD_PATTERN); } /** * Validate password with regular expression * @param password password for validation * @return true valid password, false invalid password */ public boolean validate(final String password){ matcher = pattern.matcher(password); return matcher.matches(); } } 

where 6 and 20 are minimum and maximum length for the password.

• Use a nonbacktracking expression to match the whole password first, if it has at least 8 characters (this way, there is no combinatorial explosion for long, but invalid passwords): (?>{8,})
• Use lookbehind assertions to check for presence of all required characters (AND-conditions). (?<=...)
• At least one uppercase character: (?<=\p{Lu}.*)
• At least one special character (a bit ambiguous, but let's use not-word): (?<=\W.*)
• At least one alphanumeric character (: (?<=\w.*)

Summed up:

(?>.{8,})(?<=\p{Lu}.*)(?<=\W.*)(?<=\w.*)

Best is not using regex for everything. Those requirements are very light. On CPU-wise string operations for checking the criteria/validation is much cheaper and faster than regex!

The one I found for a minimum of 8 characters, consist of at least 1 lowercase and 1 uppercase and 1 number and 1 symbols.

^((?=.*\d)(?=.*[A-Z])(?=.*\W).{8,})$