How to validate password with regular expression

Password Regular Expression Pattern



(			# Start of group
  (?=.*\d)		#   must contains one digit from 0-9
  (?=.*[a-z])		#   must contains one lowercase characters
  (?=.*[A-Z])		#   must contains one uppercase characters
  (?=.*[@#$%])		#   must contains one special symbols in the list "@#$%"
              .		#     match anything with previous condition checking
                {6,20}	#        length at least 6 characters and maximum of 20	
)			# End of group

?= – means apply the assertion condition, meaningless by itself, always work with other combination

Whole combination is means, 6 to 20 characters string with at least one digit, one upper case letter, one lower case letter and one special symbol (“@#$%”). This regular expression pattern is very useful to implement a strong and complex password.

P.S The grouping formula order is doesn’t matter.

1. Java Regular Expression Example
package com.mkyong.regex;
import java.util.regex.Matcher;
import java.util.regex.Pattern;
public class PasswordValidator{
	  private Pattern pattern;
	  private Matcher matcher;
	  private static final String PASSWORD_PATTERN = 
	  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();

2. Password that match:

1. mkyong1A@
2. mkYOn12$

3. Password that doesn’t match:

1. mY1A@ , too short, minimum 6 characters
2. mkyong12@ , uppercase characters is required
3. mkyoNg12* , special symbol “*” is not allow here
4. mkyonG$$, digit is required
5. MKYONG12$ , lower case character is required

4. Unit Test – PasswordValidator

Unit test with TestNG.
package com.mkyong.regex;
import org.testng.Assert;
import org.testng.annotations.*;
 * Password validator Testing
 * @author mkyong
public class PasswordValidatorTest {
	private PasswordValidator passwordValidator;
        public void initData(){
		passwordValidator = new PasswordValidator();
	public Object[][] ValidPasswordProvider() {
		return new Object[][]{
		   {new String[] {
			   "mkyong1A@", "mkYOn12$", 
	public Object[][] InvalidPasswordProvider() {
		return new Object[][]{
		   {new String[] {
	@Test(dataProvider = "ValidPasswordProvider")
	public void ValidPasswordTest(String[] password) {
	   for(String temp : password){
		boolean valid = passwordValidator.validate(temp);
		System.out.println("Password is valid : " + temp + " , " + valid);
		Assert.assertEquals(true, valid);
	@Test(dataProvider = "InvalidPasswordProvider", 
	public void InValidPasswordTest(String[] password) {
	   for(String temp : password){
		boolean valid = passwordValidator.validate(temp);
		System.out.println("Password is valid : " + temp + " , " + valid);
		Assert.assertEquals(false, valid);

5. Unit Test – Result

Password is valid : mkyong1A@ , true
Password is valid : mkYOn12$ , true
Password is valid : mY1A@ , false
Password is valid : mkyong12@ , false
Password is valid : mkyoNg12* , false
Password is valid : mkyonG$$ , false
Password is valid : MKYONG12$ , false
PASSED: ValidPasswordTest([Ljava.lang.String;@1d4c61c)
PASSED: InValidPasswordTest([Ljava.lang.String;@116471f)
    Tests run: 2, Failures: 0, Skips: 0
Total tests run: 2, Failures: 0, Skips: 0
Tags :

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  • Maycon

    this allow any caractere,except space, and it must to contain a letter and a number:


  • seded

    to escape the . and , from the regex

  • ih


  • raj

    i need regular expression for password validation which accepts only one character
    (from a-z) and any number of digits where password size is 8 characters.
    for ex:143h6434—> valid
    143d432y—> invalid

    in spring mvc

    thanks in advance,

  • Fernie

    Hi All,

    What if password should not include easy-to-guess string such as “love”, “happy”, “12345678”, “qwerty”, “asdfgh”, “zxcvb”. How can regular expression validate such strings?


  • Marie

    Thank you very much. Pretty helpful!

  • Anonymous


    required; min 1 lowercase letter, min 1 uppercase letter, @#$%_- special character accepting. disallow spaces, minlength 8 maxlength 20 character.

    good luck.

  • mrlami

    Dude… Awesome!

  • Anonymous

    Hi Mkyong.
    what is the pattern for gmail passwords?

  • Aniketh

    Thanks ….. works great

  • muneeb

    c program ask user to enter password of 6 character and check wether it is a strong password

  • Konrad

    I recommend: ((?=.*\d)(?=.*[a-z])(?=.*[A-Z])(?=.*[@#$%]).{6,20})([a-zA-Z\@\#\$\%\d])
    this will be block other marks

  • rudresh

    Hi its works perfect but first letter should be in character, how to add that.

    i tried adding like below in the beginning but its expecting again the capital or small letter; ex:

    1) Rudresh.12s its return false; its expects Upper case letter again
    2) rUDRESH.12s it return false; its expects Lower case letter again


    • OtaTat

      Try this one

  • shiva

    it’s accepting space

  • Raymond Ng

    Add (?!.*\\s) to disallow spaces in the password.

    • mkyong

      That’s good hack, thanks ~

  • Tomek

    3. mkyoNg12* , special symbol “*” is not allow here

    true, but try this:
    Ng1#**** – allowed!

    (in fact there could be ANY char in place of ‘*';

    If you would like to limit chars to only [a-zA-Z0-9@#$%] use:


  • Haider M Rizvi

    This post helped me. Thanks.

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  • Lincoln Baxter, III

    Hey! Great example – I’d like to suggest a slight adaptation, however. While it is tempting to use a single regular expression for this, I think that there are good reasons to actually split up the regex into multiple checks. Performance is not usually a concern with password checking, so invoking a few more regex calls isn’t really a big deal, like so:

  • chris k

    Thanks for this! Saved me a lot of time. Much appreciated Mkyong!

  • Jonas Grimsgaard

    Thank you, you saved me ALOT of time ?

  • Belen Kotow

    yeah, you are right. this is a very good articles.i have learned so many things from

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  • Jeremiah

    Using this string as test data: “X@CpJ[8~”

    It would return true, even though the characters ‘[‘ and ‘~’ are not allowed.

    • Victor

      According to the regex these characters are allowed, but not required.

  • John

    This regEx fails for April123
    Why is that so?

    According to pattern it should not pass right?

    • Satish Motwani

      Hello John,
      Atleast one special character out of [@#$%] must be present.

  • Mo Fielding

    Thanks! Very helpful. I slept through the regex stuff in class… :-(

  • Alex

    Good night,

    I’m not able to pass parameters to the regular expression for example:
    ((?=.*\\d{3}) == Change de number 3 to a variable .

    Thanks a lot,

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