Swift - Complex Regular Expression on Password Field [duplicate] - ios

I want a regular expression to check that:
A password contains at least eight characters, including at least one number and includes both lower and uppercase letters and special characters, for example #, ?, !.
It cannot be your old password or contain your username, "password", or "websitename"
And here is my validation expression which is for eight characters including one uppercase letter, one lowercase letter, and one number or special character.
(?=^.{8,}$)((?=.*\d)|(?=.*\W+))(?![.\n])(?=.*[A-Z])(?=.*[a-z]).*$"
How can I write it for a password must be eight characters including one uppercase letter, one special character and alphanumeric characters?

Minimum eight characters, at least one letter and one number:
"^(?=.*[A-Za-z])(?=.*\d)[A-Za-z\d]{8,}$"
Minimum eight characters, at least one letter, one number and one special character:
"^(?=.*[A-Za-z])(?=.*\d)(?=.*[#$!%*#?&])[A-Za-z\d#$!%*#?&]{8,}$"
Minimum eight characters, at least one uppercase letter, one lowercase letter and one number:
"^(?=.*[a-z])(?=.*[A-Z])(?=.*\d)[a-zA-Z\d]{8,}$"
Minimum eight characters, at least one uppercase letter, one lowercase letter, one number and one special character:
"^(?=.*[a-z])(?=.*[A-Z])(?=.*\d)(?=.*[#$!%*?&])[A-Za-z\d#$!%*?&]{8,}$"
Minimum eight and maximum 10 characters, at least one uppercase letter, one lowercase letter, one number and one special character:
"^(?=.*[a-z])(?=.*[A-Z])(?=.*\d)(?=.*[#$!%*?&])[A-Za-z\d#$!%*?&]{8,10}$"

You may use this regex with multiple lookahead assertions (conditions):
^(?=.*?[A-Z])(?=.*?[a-z])(?=.*?[0-9])(?=.*?[#?!#$%^&*-]).{8,}$
This regex will enforce these rules:
At least one upper case English letter, (?=.*?[A-Z])
At least one lower case English letter, (?=.*?[a-z])
At least one digit, (?=.*?[0-9])
At least one special character, (?=.*?[#?!#$%^&*-])
Minimum eight in length .{8,} (with the anchors)

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 eight characters OR anything with no numbers OR anything with no uppercase OR or anything with no lowercase OR anything with no special characters.
So:
^(.{0,7}|[^0-9]*|[^A-Z]*|[^a-z]*|[a-zA-Z0-9]*)$
If anything matches that, then it's an invalid password.

Use the following Regex to satisfy the below conditions:
Conditions:
Min 1 uppercase letter.
Min 1 lowercase letter.
Min 1 special character.
Min 1 number.
Min 8 characters.
Max 30 characters.
Regex:
/^(?=.*[a-z])(?=.*[A-Z])(?=.*\d)(?=.*[#$#!%&*?])[A-Za-z\d#$#!%&*?]{8,30}$/

Just a small improvement for #anubhava's answer: Since special character are limited to the ones in the keyboard, use this for any special character:
^(?=.*?[A-Z])(?=(.*[a-z]){1,})(?=(.*[\d]){1,})(?=(.*[\W]){1,})(?!.*\s).{8,}$
This regex will enforce these rules:
At least one upper case English letter
At least one lower case English letter
At least one digit
At least one special character
Minimum eight in length

I had some difficulty following the most popular answer for my circumstances. For example, my validation was failing with characters such as ; or [. I was not interested in white-listing my special characters, so I instead leveraged [^\w\s] as a test - simply put - match non word characters (including numeric) and non white space characters. To summarize, here is what worked for me...
at least 8 characters
at least 1 numeric character
at least 1 lowercase letter
at least 1 uppercase letter
at least 1 special character
/^(?=.*?[A-Z])(?=.*?[a-z])(?=.*?[0-9])(?=.*?[^\w\s]).{8,}$/
JSFiddle Link - simple demo covering various cases

 
✅ The following 4 regex patterns can help you to write almost any password validation
 
 
Pattern 1:
 
Password must contain one digit from 1 to 9, one lowercase letter, one uppercase letter, one special character, no space, and it must be 8-16 characters long.
/^(?=.*[0-9])(?=.*[a-z])(?=.*[A-Z])(?=.*\W)(?!.* ).{8,16}$/
 
Explanation:
 
(?=.*[0-9]) means that the password must contain a single digit from 1 to 9.
 
(?=.*[a-z]) means that the password must contain one lowercase letter.
 
(?=.*[A-Z]) means that the password must contain one uppercase letter.
 
(?=.*\W) means that the password must contain one special character.
 
.{8,16} means that the password must be 8-16 characters long. We must use this at the end of the regex, just before the $ symbol.
 
What are ^ and $:
 
^ indicates the beginning of the string. $ indicates the end of the string.
If we don't use these ^ & $, the regex will not be able to determine the maximum length of the password. In the above example, we have a condition that the password can't be longer than 16 characters, to make that condition work, we have used these ^ & $
 
Remove maximum length restriction:
 
Instead of .{8,16}, if we used .{8,}, it would mean that the password must be at least 8 characters long. So, there will not be any condition for checking the maximum length of the password.
 
Don't accept any number(digit):
 
Instead of (?=.*[0-9]), if we used (?!.*[0-9]), it would mean that the password must not contain any digit from 1-9 (Difference with the (?=.*[0-9]) is the use of ! instead of =)
 
Don't accept any spcecial character:
 
Instead of (?=.*\W), if we used (?!.*\W), it would mean that the password must not contain any special characters (The difference with the (?=.*\W) is the use of ! instead of =)
 
Alternative Syntax for number(digit):
 
Instead of (?=.*[0-9]), we could have used (?=.*\d). (?=.*\d) also means that the password must contain a single digit from 1 to 9.
 
 
Pattern 2:
 
Password must contain one digit from 1 to 9, one lowercase letter, one uppercase letter, one underscore but no other special character, no space and it must be 8-16 characters long.
/^(?=.*[0-9])(?=.*[a-z])(?=.*[A-Z])(?=.*_)(?!.*\W)(?!.* ).{8,16}$/
 
Difference with the Pattern 1
 
Here, we have used (?=.*_) which wasn't on the Pattern 1.
 
(?=.*_)(?!.*\W) means that the password must contain an underscore but can not contain any other special character.
 
Pattern 3:
 
Password must contain one digit from 1 to 9, one lowercase letter, one uppercase letter, one underscore, no space and it must be 8-16 characters long. Usage of any other special character other than underscore is optional.
/^(?=.*[0-9])(?=.*[a-z])(?=.*[A-Z])(?=.*_)(?!.* ).{8,16}$/
 
Difference with the Pattern 2
 
Here, we have not used (?!.*\W) what was on the Pattern 2.
 
But it still has the (?=.*_)
 
By just removing the (?!.*\W), special characters have become optional. Now, one underscore is required but any other special character can be used or not as it's optional.
 
Pattern 4:
 
Password must contain one digit from 1 to 9, one lowercase letter, one uppercase letter, and one underscore, and it must be 8-16 characters long. Usage of any other special character and usage of space is optional.
/^(?=.*[0-9])(?=.*[a-z])(?=.*[A-Z]).{8,16}$/
 
Difference with the Pattern 3
 
Here, we have not used (?=.*_) & (?!.* ) which was on the Pattern 3.
 
By removing (?=.*_), it's no longer mandatory to pass one underscore. Now, passing special characters is optional.
 
By removing the (?!.* ), usage of space has become optional too.

I would reply to Peter Mortensen, but I don't have enough reputation.
His expressions are perfect for each of the specified minimum requirements. The problem with his expressions that don't require special characters is that they also don't ALLOW special characters, so they also enforce maximum requirements, which I don't believe the OP requested. Normally you want to allow your users to make their password as strong as they want; why restrict strong passwords?
So, his "minimum eight characters, at least one letter and one number" expression:
^(?=.*[A-Za-z])(?=.*\d)[A-Za-z\d]{8,}$
achieves the minimum requirement, but the remaining characters can only be letter and numbers. To allow (but not require) special characters, you should use something like:
^(?=.*[A-Za-z])(?=.*\d).{8,}$ to allow any characters
or
^(?=.*[A-Za-z])(?=.*\d)[A-Za-z\d$#$!%*#?&]{8,}$ to allow specific special characters
Likewise, "minimum eight characters, at least one uppercase letter, one lowercase letter and one number:"
^(?=.*[a-z])(?=.*[A-Z])(?=.*\d)[a-zA-Z\d]{8,}$
meets that minimum requirement, but only allows letters and numbers. Use:
^(?=.*[a-z])(?=.*[A-Z])(?=.*\d).{8,}$ to allow any characters
or
^(?=.*[a-z])(?=.*[A-Z])(?=.*\d)[A-Za-z\d$#$!%*?&]{8,} to allow specific special characters.

A more "generic" version(?), allowing none English letters as special characters.
^(?=\S*[a-z])(?=\S*[A-Z])(?=\S*\d)(?=\S*[^\w\s])\S{8,}$
var pwdList = [
'##V4-\3Z`zTzM{>k',
'12qw!"QW12',
'123qweASD!"#',
'1qA!"#$%&',
'Günther32',
'123456789',
'qweASD123',
'qweqQWEQWEqw',
'12qwAS!'
],
re = /^(?=\S*[a-z])(?=\S*[A-Z])(?=\S*\d)(?=\S*[^\w\s])\S{8,}$/;
pwdList.forEach(function (pw) {
document.write('<span style="color:'+ (re.test(pw) ? 'green':'red') + '">' + pw + '</span><br/>');
});

Import the JavaScript file jquery.validate.min.js.
You can use this method:
$.validator.addMethod("pwcheck", function (value) {
return /[\#\#\$\%\^\&\*\(\)\_\+\!]/.test(value) && /[a-z]/.test(value) && /[0-9]/.test(value) && /[A-Z]/.test(value)
});
At least one upper case English letter
At least one lower case English letter
At least one digit
At least one special character

For standard password requirements I found this to be useful:
At least 1 alphabet
At least 1 digit
Contains no space
Optional special characters e.g. #$!%*#?&^_-
Minimum 8 characters long
/^(?=.*[A-Za-z])(?=.*\d)[A-Za-z\d#$!%*#?&^_-]{8,}$/
You can also set the upper limit for example {8,32} up to 32 characters long.

Try this one:
Minimum six characters
At least one uppercase character
At least one lowercase character
At least one special character
Expression:
"/^(?=.*[a-z])(?=.*[A-Z])(?=.*\d)(?=.*[$#$!%*?&.])[A-Za-z\d$#$!%*?&.]{6, 20}/"
Optional Special Characters:
At least one special character
At least one number
Special characters are optional
Minimum six characters and maximum 16 characters
Expression:
"/^(?=.*\d)(?=.*[a-zA-Z]).{6,20}$/"
If the min and max condition is not required then remove .{6, 16}
6 is minimum character limit
20 is maximum character limit
?= means match expression

This worked for me:
^(?=.*[0-9])(?=.*[a-z])(?=.*[A-Z])(?=.*[#$!%*?&])([a-zA-Z0-9#$!%*?&]{8,})$
At least 8 characters long;
One lowercase, one uppercase, one number and one special character;
No whitespaces.

Not directly answering the question, but does it really have to be a regex?
I used to do lots of Perl, and got used to solving problems with regexes. However, when they get more complicated with all the look-aheads and other quirks, you need to write dozens of unit tests to kill all those little bugs.
Furthermore, a regex is typically a few times slower than an imperative or a functional solution.
For example, the following (not very FP) Scala function solves the original question about three times faster than the regex of the most popular answer. What it does is also so clear that you don't need a unit test at all:
def validatePassword(password: String): Boolean = {
if (password.length < 8)
return false
var lower = false
var upper = false
var numbers = false
var special = false
password.foreach { c =>
if (c.isDigit) numbers = true
else if (c.isLower) lower = true
else if (c.isUpper) upper = true
else special = true
}
lower && upper && numbers && special
}

For a more strict validation where the following is required:
At least One Upper Case Character
At least one Lower Case character
At least one digit
At least one symbol/special character #$!%*#?&^_-
Minimum 8 characters/digits
Regex:
/(?=.*\d)(?=.*[a-z])(?=.*[A-Z])(?=.*[#$!%*#?&^_-]).{8,}/
I hope it helps someone with a more stringent.

What about considering the following regex solution:
^(?=.*[\w])(?=.*[\W])[\w\W]{8,}$
Which validates the following:
At least one lowercase
At least one uppercase
At least one digit
At least one special character
At least it should have 8 characters long.
Check it out working at the following link https://regex101.com/r/qPmC06/4/

^(?=.*[A-Z])(?=.*[a-z])(?=.*[0-9])(?=.*[!##$%^&*()_+,.\\\/;':"-]).{8,}$

Another option is to make use of contrast in the lookahead assertions using a negated character class, optionally matching any character except that is listed before matching the character that should be matched.
^(?=[^A-Z\n]*[A-Z])(?=[^a-z\n]*[a-z])(?=[^0-9\n]*[0-9])(?=[^#?!#$%^&*\n-]*[#?!#$%^&*-]).{8,}$
See a regex demo
In parts, the pattern matches:
^ Start of string
(?=[^A-Z\n]*[A-Z]) Positive lookahead, assert 0+ times any char except A-Z or a newline. Then match a char A-Z
(?=[^a-z\n]*[a-z]) The same approach for a char a-z
(?=[^0-9\n]*[0-9]) The same approach for a digit 0-9
(?=[^#?!#$%^&*\n-]*[#?!#$%^&*-]) The same approach for a char that you would consider special
.{8,} Match 8 or more times any character except a newline
$ End of string
Notes
A dot can also match a space. If you do not want to allow matching a space, then .{8,} can be changed to \S{8,} to match 8 or more non whitespace characters
Using either . or \S can match more characters than are specified in the lookahead assertions. If you only want to match the characters that are used in the assertions, you can change .{8,} to match only the allowed characters [#?!#$%^&*A-Za-z0-9-]{8,} using a character class
const regex = /^(?=[^A-Z\n]*[A-Z])(?=[^a-z\n]*[a-z])(?=[^0-9\n]*[0-9])(?=[^#?!#$%^&*\n-]*[#?!#$%^&*-]).{8,}$/;
[
"abcA1#!A",
"#!asdfSFD1;",
"# a f F1 ;",
"1111111111",
"aaaaaaaa",
"11111111",
"AAAAAAAA",
"########",
"aA1#"
].forEach(s =>
console.log(regex.test(s) ? `Match --> ${s}` : `No match --> ${s}`)
);

/^(?=.*[A-Z])(?=.*[a-z])(?=.*[0-9]).*$/
this the simple way to use it while validate atleast 1 uppercase 1 lowercase and 1 number
and this is the example while I use in express validation
check('password')
.notEmpty()
.withMessage('Password cannot be null')
.bail()
.isLength({ min: 6 })
.withMessage('Password must be at least 6 characters')
.bail()
.matches(/^(?=.*[A-Z])(?=.*[a-z])(?=.*[0-9]).*$/)
.withMessage(
'Must have atleast 1 uppercase, 1 lowercase letter and 1 number'
),

Testing this one in 2020:
^(?=.*[a-z])(?=.*[A-Z])(?=.*\d)(?=.*[#$!%*?&])[A-Za-z\d#$!%*?&]{8,}$
Verify yourself
const regex = /^(?=.*[a-z])(?=.*[A-Z])(?=.*\d)(?=.*[#$!%*?&])[A-Za-z\d#$!%*?&]{8,}$/;
const str = `some12*Nuts`;
let m;
if ((m = regex.exec(str)) !== null) {
// The result can be accessed through the `m`-variable.
m.forEach((match, groupIndex) => {
console.log(`Found match, group ${groupIndex}: ${match}`);
});
}

#ClasG has already suggested:
^(?=\S*[a-z])(?=\S*[A-Z])(?=\S*\d)(?=\S*[^\w\s])\S{8,}$
but it does not accept _(underscore) as a special character (eg. Aa12345_).
An improved one is:
^(?=\S*[a-z])(?=\S*[A-Z])(?=\S*\d)(?=\S*([^\w\s]|[_]))\S{8,}$

I've found many problems here, so I made my own.
Here it is in all it's glory, with tests:
^(?=.*[a-z])(?=.*[A-Z])(?=.*\d)(?=.*([^a-zA-Z\d\s])).{9,}$
https://regex101.com/r/DCRR65/4/tests
Things to look out for:
doesn't use \w because that includes _, which I'm testing for.
I've had lots of troubles matching symbols, without matching the end of the line.
Doesn't specify symbols specifically, this is also because different locales may have different symbols on their keyboards that they may want to use.

Demo:
function password_check() {
pass = document.getElementById("password").value;
console.log(pass);
regex = /^(?=.*[a-z])(?=.*[A-Z])(?=.*\d)(?=.*[#$!%*?&])[A-Za-z\d#$!%*?&]{8,}$/;
if (regex.exec(pass) == null) {
alert('invalid password!')
}
else {
console.log("valid");
}
}
<input type="text" id="password" value="Sample#1">
<input type="button" id="submit" onclick="password_check()" value="submit">

var strongRegex = new RegExp("^(?=.*[a-z])(?=.*[A-Z])(?=.*[0-9])(?=.*[!##\$%\^&\*])(?=.{8,})");
var mediumRegex = new RegExp("^(((?=.*[a-z])(?=.*[A-Z]))|((?=.*[a-z])(?=.*[0-9]))|((?=.*[A-Z])(?=.*[0-9])))(?=.{6,})");
Best For javascript

Keep it simple stupid:
This should do the trick for you, always.
Regex: ^(.{0,7}|[^a-z]{1,}|[^A-Z]{1,}|[^\d]{1,}|[^\W]{1,})$|[\s]
If your password matches the regex above, it is invalid.
If there's no match, your password is valid and contains has at least 8 characters, one upper case letter, one lower case letter and one symbol or special character. And it also contains no spaces, tabs or line breaks.
Breakdown of Regex
.{0,7} - matches if password has between 0 to 7 characters.
[^a-z]{1,} - matches if no lower case is found
[^A-Z]{1,} - matches if no upper case is found
[^\d]{1,} - matches if no number (between [0-9]) is found
[\s] - matches if a white space, tab or line break is found.
With this approach there's no limit or restriction in terms of symbols allowed. If you want to limit to few symbols allowable, just change [^\W] with [^YourSymbols].

(?=.*[0-9])(?=.*[a-z])(?=.*[A-Z])(?=.*[##$%^&+-]).{6}

According to your need this pattern should work just fine. Try this,
^(?=(.*\d){1})(.*\S)(?=.*[a-zA-Z\S])[0-9a-zA-Z\S]{8,}
Just create a string variable, assign the pattern, and create a boolean method which returns true if the pattern is correct, else false.
Sample:
String pattern = "^(?=(.*\d){1})(.*\S)(?=.*[a-zA-Z\S])[0-9a-zA-Z\S]{8,}";
String password_string = "Type the password here"
private boolean isValidPassword(String password_string) {
return password_string.matches(Constants.passwordPattern);
}

Use the following Regex to satisfy the below conditions:
Conditions: 1] Min 1 special character.
2] Min 1 number.
3] Min 8 characters or More
Regex: ^(?=.*\d)(?=.*[#$#!%&*?])[A-Za-z\d#$#!%&*?]{8,}$
Can Test Online: https://regex101.com

Just we can do this by using HTML5.
Use below code in pattern attribute,
pattern="(?=^.{8,}$)((?=.*\d)(?=.*\W+))(?![.\n])(?=.*[A-Z])(?=.*[a-z]).*$"
It will work perfectly.

You can use the below regular expression pattern to check the password whether it matches your expectations or not.
((?=.*\\d)(?=.*[a-z])(?=.*[A-Z])(?=.*[~!##$%^&*()]).{8,20})

Related

Check password length with regex [duplicate]

I want a regular expression to check that:
A password contains at least eight characters, including at least one number and includes both lower and uppercase letters and special characters, for example #, ?, !.
It cannot be your old password or contain your username, "password", or "websitename"
And here is my validation expression which is for eight characters including one uppercase letter, one lowercase letter, and one number or special character.
(?=^.{8,}$)((?=.*\d)|(?=.*\W+))(?![.\n])(?=.*[A-Z])(?=.*[a-z]).*$"
How can I write it for a password must be eight characters including one uppercase letter, one special character and alphanumeric characters?
Minimum eight characters, at least one letter and one number:
"^(?=.*[A-Za-z])(?=.*\d)[A-Za-z\d]{8,}$"
Minimum eight characters, at least one letter, one number and one special character:
"^(?=.*[A-Za-z])(?=.*\d)(?=.*[#$!%*#?&])[A-Za-z\d#$!%*#?&]{8,}$"
Minimum eight characters, at least one uppercase letter, one lowercase letter and one number:
"^(?=.*[a-z])(?=.*[A-Z])(?=.*\d)[a-zA-Z\d]{8,}$"
Minimum eight characters, at least one uppercase letter, one lowercase letter, one number and one special character:
"^(?=.*[a-z])(?=.*[A-Z])(?=.*\d)(?=.*[#$!%*?&])[A-Za-z\d#$!%*?&]{8,}$"
Minimum eight and maximum 10 characters, at least one uppercase letter, one lowercase letter, one number and one special character:
"^(?=.*[a-z])(?=.*[A-Z])(?=.*\d)(?=.*[#$!%*?&])[A-Za-z\d#$!%*?&]{8,10}$"
You may use this regex with multiple lookahead assertions (conditions):
^(?=.*?[A-Z])(?=.*?[a-z])(?=.*?[0-9])(?=.*?[#?!#$%^&*-]).{8,}$
This regex will enforce these rules:
At least one upper case English letter, (?=.*?[A-Z])
At least one lower case English letter, (?=.*?[a-z])
At least one digit, (?=.*?[0-9])
At least one special character, (?=.*?[#?!#$%^&*-])
Minimum eight in length .{8,} (with the anchors)
Regular expressions don't have an AND operator, so it's pretty hard to write a regex that matches valid passwords, when validity is defined by something AND something else AND something else...
But, regular expressions do have an OR operator, so just apply DeMorgan's theorem, and write a regex that matches invalid passwords:
Anything with less than eight characters OR anything with no numbers OR anything with no uppercase OR or anything with no lowercase OR anything with no special characters.
So:
^(.{0,7}|[^0-9]*|[^A-Z]*|[^a-z]*|[a-zA-Z0-9]*)$
If anything matches that, then it's an invalid password.
Use the following Regex to satisfy the below conditions:
Conditions:
Min 1 uppercase letter.
Min 1 lowercase letter.
Min 1 special character.
Min 1 number.
Min 8 characters.
Max 30 characters.
Regex:
/^(?=.*[a-z])(?=.*[A-Z])(?=.*\d)(?=.*[#$#!%&*?])[A-Za-z\d#$#!%&*?]{8,30}$/
Just a small improvement for #anubhava's answer: Since special character are limited to the ones in the keyboard, use this for any special character:
^(?=.*?[A-Z])(?=(.*[a-z]){1,})(?=(.*[\d]){1,})(?=(.*[\W]){1,})(?!.*\s).{8,}$
This regex will enforce these rules:
At least one upper case English letter
At least one lower case English letter
At least one digit
At least one special character
Minimum eight in length
I had some difficulty following the most popular answer for my circumstances. For example, my validation was failing with characters such as ; or [. I was not interested in white-listing my special characters, so I instead leveraged [^\w\s] as a test - simply put - match non word characters (including numeric) and non white space characters. To summarize, here is what worked for me...
at least 8 characters
at least 1 numeric character
at least 1 lowercase letter
at least 1 uppercase letter
at least 1 special character
/^(?=.*?[A-Z])(?=.*?[a-z])(?=.*?[0-9])(?=.*?[^\w\s]).{8,}$/
JSFiddle Link - simple demo covering various cases
 
✅ The following 4 regex patterns can help you to write almost any password validation
 
 
Pattern 1:
 
Password must contain one digit from 1 to 9, one lowercase letter, one uppercase letter, one special character, no space, and it must be 8-16 characters long.
/^(?=.*[0-9])(?=.*[a-z])(?=.*[A-Z])(?=.*\W)(?!.* ).{8,16}$/
 
Explanation:
 
(?=.*[0-9]) means that the password must contain a single digit from 1 to 9.
 
(?=.*[a-z]) means that the password must contain one lowercase letter.
 
(?=.*[A-Z]) means that the password must contain one uppercase letter.
 
(?=.*\W) means that the password must contain one special character.
 
.{8,16} means that the password must be 8-16 characters long. We must use this at the end of the regex, just before the $ symbol.
 
What are ^ and $:
 
^ indicates the beginning of the string. $ indicates the end of the string.
If we don't use these ^ & $, the regex will not be able to determine the maximum length of the password. In the above example, we have a condition that the password can't be longer than 16 characters, to make that condition work, we have used these ^ & $
 
Remove maximum length restriction:
 
Instead of .{8,16}, if we used .{8,}, it would mean that the password must be at least 8 characters long. So, there will not be any condition for checking the maximum length of the password.
 
Don't accept any number(digit):
 
Instead of (?=.*[0-9]), if we used (?!.*[0-9]), it would mean that the password must not contain any digit from 1-9 (Difference with the (?=.*[0-9]) is the use of ! instead of =)
 
Don't accept any spcecial character:
 
Instead of (?=.*\W), if we used (?!.*\W), it would mean that the password must not contain any special characters (The difference with the (?=.*\W) is the use of ! instead of =)
 
Alternative Syntax for number(digit):
 
Instead of (?=.*[0-9]), we could have used (?=.*\d). (?=.*\d) also means that the password must contain a single digit from 1 to 9.
 
 
Pattern 2:
 
Password must contain one digit from 1 to 9, one lowercase letter, one uppercase letter, one underscore but no other special character, no space and it must be 8-16 characters long.
/^(?=.*[0-9])(?=.*[a-z])(?=.*[A-Z])(?=.*_)(?!.*\W)(?!.* ).{8,16}$/
 
Difference with the Pattern 1
 
Here, we have used (?=.*_) which wasn't on the Pattern 1.
 
(?=.*_)(?!.*\W) means that the password must contain an underscore but can not contain any other special character.
 
Pattern 3:
 
Password must contain one digit from 1 to 9, one lowercase letter, one uppercase letter, one underscore, no space and it must be 8-16 characters long. Usage of any other special character other than underscore is optional.
/^(?=.*[0-9])(?=.*[a-z])(?=.*[A-Z])(?=.*_)(?!.* ).{8,16}$/
 
Difference with the Pattern 2
 
Here, we have not used (?!.*\W) what was on the Pattern 2.
 
But it still has the (?=.*_)
 
By just removing the (?!.*\W), special characters have become optional. Now, one underscore is required but any other special character can be used or not as it's optional.
 
Pattern 4:
 
Password must contain one digit from 1 to 9, one lowercase letter, one uppercase letter, and one underscore, and it must be 8-16 characters long. Usage of any other special character and usage of space is optional.
/^(?=.*[0-9])(?=.*[a-z])(?=.*[A-Z]).{8,16}$/
 
Difference with the Pattern 3
 
Here, we have not used (?=.*_) & (?!.* ) which was on the Pattern 3.
 
By removing (?=.*_), it's no longer mandatory to pass one underscore. Now, passing special characters is optional.
 
By removing the (?!.* ), usage of space has become optional too.
I would reply to Peter Mortensen, but I don't have enough reputation.
His expressions are perfect for each of the specified minimum requirements. The problem with his expressions that don't require special characters is that they also don't ALLOW special characters, so they also enforce maximum requirements, which I don't believe the OP requested. Normally you want to allow your users to make their password as strong as they want; why restrict strong passwords?
So, his "minimum eight characters, at least one letter and one number" expression:
^(?=.*[A-Za-z])(?=.*\d)[A-Za-z\d]{8,}$
achieves the minimum requirement, but the remaining characters can only be letter and numbers. To allow (but not require) special characters, you should use something like:
^(?=.*[A-Za-z])(?=.*\d).{8,}$ to allow any characters
or
^(?=.*[A-Za-z])(?=.*\d)[A-Za-z\d$#$!%*#?&]{8,}$ to allow specific special characters
Likewise, "minimum eight characters, at least one uppercase letter, one lowercase letter and one number:"
^(?=.*[a-z])(?=.*[A-Z])(?=.*\d)[a-zA-Z\d]{8,}$
meets that minimum requirement, but only allows letters and numbers. Use:
^(?=.*[a-z])(?=.*[A-Z])(?=.*\d).{8,}$ to allow any characters
or
^(?=.*[a-z])(?=.*[A-Z])(?=.*\d)[A-Za-z\d$#$!%*?&]{8,} to allow specific special characters.
A more "generic" version(?), allowing none English letters as special characters.
^(?=\S*[a-z])(?=\S*[A-Z])(?=\S*\d)(?=\S*[^\w\s])\S{8,}$
var pwdList = [
'##V4-\3Z`zTzM{>k',
'12qw!"QW12',
'123qweASD!"#',
'1qA!"#$%&',
'Günther32',
'123456789',
'qweASD123',
'qweqQWEQWEqw',
'12qwAS!'
],
re = /^(?=\S*[a-z])(?=\S*[A-Z])(?=\S*\d)(?=\S*[^\w\s])\S{8,}$/;
pwdList.forEach(function (pw) {
document.write('<span style="color:'+ (re.test(pw) ? 'green':'red') + '">' + pw + '</span><br/>');
});
Import the JavaScript file jquery.validate.min.js.
You can use this method:
$.validator.addMethod("pwcheck", function (value) {
return /[\#\#\$\%\^\&\*\(\)\_\+\!]/.test(value) && /[a-z]/.test(value) && /[0-9]/.test(value) && /[A-Z]/.test(value)
});
At least one upper case English letter
At least one lower case English letter
At least one digit
At least one special character
For standard password requirements I found this to be useful:
At least 1 alphabet
At least 1 digit
Contains no space
Optional special characters e.g. #$!%*#?&^_-
Minimum 8 characters long
/^(?=.*[A-Za-z])(?=.*\d)[A-Za-z\d#$!%*#?&^_-]{8,}$/
You can also set the upper limit for example {8,32} up to 32 characters long.
Try this one:
Minimum six characters
At least one uppercase character
At least one lowercase character
At least one special character
Expression:
"/^(?=.*[a-z])(?=.*[A-Z])(?=.*\d)(?=.*[$#$!%*?&.])[A-Za-z\d$#$!%*?&.]{6, 20}/"
Optional Special Characters:
At least one special character
At least one number
Special characters are optional
Minimum six characters and maximum 16 characters
Expression:
"/^(?=.*\d)(?=.*[a-zA-Z]).{6,20}$/"
If the min and max condition is not required then remove .{6, 16}
6 is minimum character limit
20 is maximum character limit
?= means match expression
This worked for me:
^(?=.*[0-9])(?=.*[a-z])(?=.*[A-Z])(?=.*[#$!%*?&])([a-zA-Z0-9#$!%*?&]{8,})$
At least 8 characters long;
One lowercase, one uppercase, one number and one special character;
No whitespaces.
Not directly answering the question, but does it really have to be a regex?
I used to do lots of Perl, and got used to solving problems with regexes. However, when they get more complicated with all the look-aheads and other quirks, you need to write dozens of unit tests to kill all those little bugs.
Furthermore, a regex is typically a few times slower than an imperative or a functional solution.
For example, the following (not very FP) Scala function solves the original question about three times faster than the regex of the most popular answer. What it does is also so clear that you don't need a unit test at all:
def validatePassword(password: String): Boolean = {
if (password.length < 8)
return false
var lower = false
var upper = false
var numbers = false
var special = false
password.foreach { c =>
if (c.isDigit) numbers = true
else if (c.isLower) lower = true
else if (c.isUpper) upper = true
else special = true
}
lower && upper && numbers && special
}
For a more strict validation where the following is required:
At least One Upper Case Character
At least one Lower Case character
At least one digit
At least one symbol/special character #$!%*#?&^_-
Minimum 8 characters/digits
Regex:
/(?=.*\d)(?=.*[a-z])(?=.*[A-Z])(?=.*[#$!%*#?&^_-]).{8,}/
I hope it helps someone with a more stringent.
What about considering the following regex solution:
^(?=.*[\w])(?=.*[\W])[\w\W]{8,}$
Which validates the following:
At least one lowercase
At least one uppercase
At least one digit
At least one special character
At least it should have 8 characters long.
Check it out working at the following link https://regex101.com/r/qPmC06/4/
^(?=.*[A-Z])(?=.*[a-z])(?=.*[0-9])(?=.*[!##$%^&*()_+,.\\\/;':"-]).{8,}$
Another option is to make use of contrast in the lookahead assertions using a negated character class, optionally matching any character except that is listed before matching the character that should be matched.
^(?=[^A-Z\n]*[A-Z])(?=[^a-z\n]*[a-z])(?=[^0-9\n]*[0-9])(?=[^#?!#$%^&*\n-]*[#?!#$%^&*-]).{8,}$
See a regex demo
In parts, the pattern matches:
^ Start of string
(?=[^A-Z\n]*[A-Z]) Positive lookahead, assert 0+ times any char except A-Z or a newline. Then match a char A-Z
(?=[^a-z\n]*[a-z]) The same approach for a char a-z
(?=[^0-9\n]*[0-9]) The same approach for a digit 0-9
(?=[^#?!#$%^&*\n-]*[#?!#$%^&*-]) The same approach for a char that you would consider special
.{8,} Match 8 or more times any character except a newline
$ End of string
Notes
A dot can also match a space. If you do not want to allow matching a space, then .{8,} can be changed to \S{8,} to match 8 or more non whitespace characters
Using either . or \S can match more characters than are specified in the lookahead assertions. If you only want to match the characters that are used in the assertions, you can change .{8,} to match only the allowed characters [#?!#$%^&*A-Za-z0-9-]{8,} using a character class
const regex = /^(?=[^A-Z\n]*[A-Z])(?=[^a-z\n]*[a-z])(?=[^0-9\n]*[0-9])(?=[^#?!#$%^&*\n-]*[#?!#$%^&*-]).{8,}$/;
[
"abcA1#!A",
"#!asdfSFD1;",
"# a f F1 ;",
"1111111111",
"aaaaaaaa",
"11111111",
"AAAAAAAA",
"########",
"aA1#"
].forEach(s =>
console.log(regex.test(s) ? `Match --> ${s}` : `No match --> ${s}`)
);
/^(?=.*[A-Z])(?=.*[a-z])(?=.*[0-9]).*$/
this the simple way to use it while validate atleast 1 uppercase 1 lowercase and 1 number
and this is the example while I use in express validation
check('password')
.notEmpty()
.withMessage('Password cannot be null')
.bail()
.isLength({ min: 6 })
.withMessage('Password must be at least 6 characters')
.bail()
.matches(/^(?=.*[A-Z])(?=.*[a-z])(?=.*[0-9]).*$/)
.withMessage(
'Must have atleast 1 uppercase, 1 lowercase letter and 1 number'
),
Testing this one in 2020:
^(?=.*[a-z])(?=.*[A-Z])(?=.*\d)(?=.*[#$!%*?&])[A-Za-z\d#$!%*?&]{8,}$
Verify yourself
const regex = /^(?=.*[a-z])(?=.*[A-Z])(?=.*\d)(?=.*[#$!%*?&])[A-Za-z\d#$!%*?&]{8,}$/;
const str = `some12*Nuts`;
let m;
if ((m = regex.exec(str)) !== null) {
// The result can be accessed through the `m`-variable.
m.forEach((match, groupIndex) => {
console.log(`Found match, group ${groupIndex}: ${match}`);
});
}
#ClasG has already suggested:
^(?=\S*[a-z])(?=\S*[A-Z])(?=\S*\d)(?=\S*[^\w\s])\S{8,}$
but it does not accept _(underscore) as a special character (eg. Aa12345_).
An improved one is:
^(?=\S*[a-z])(?=\S*[A-Z])(?=\S*\d)(?=\S*([^\w\s]|[_]))\S{8,}$
I've found many problems here, so I made my own.
Here it is in all it's glory, with tests:
^(?=.*[a-z])(?=.*[A-Z])(?=.*\d)(?=.*([^a-zA-Z\d\s])).{9,}$
https://regex101.com/r/DCRR65/4/tests
Things to look out for:
doesn't use \w because that includes _, which I'm testing for.
I've had lots of troubles matching symbols, without matching the end of the line.
Doesn't specify symbols specifically, this is also because different locales may have different symbols on their keyboards that they may want to use.
Demo:
function password_check() {
pass = document.getElementById("password").value;
console.log(pass);
regex = /^(?=.*[a-z])(?=.*[A-Z])(?=.*\d)(?=.*[#$!%*?&])[A-Za-z\d#$!%*?&]{8,}$/;
if (regex.exec(pass) == null) {
alert('invalid password!')
}
else {
console.log("valid");
}
}
<input type="text" id="password" value="Sample#1">
<input type="button" id="submit" onclick="password_check()" value="submit">
var strongRegex = new RegExp("^(?=.*[a-z])(?=.*[A-Z])(?=.*[0-9])(?=.*[!##\$%\^&\*])(?=.{8,})");
var mediumRegex = new RegExp("^(((?=.*[a-z])(?=.*[A-Z]))|((?=.*[a-z])(?=.*[0-9]))|((?=.*[A-Z])(?=.*[0-9])))(?=.{6,})");
Best For javascript
Keep it simple stupid:
This should do the trick for you, always.
Regex: ^(.{0,7}|[^a-z]{1,}|[^A-Z]{1,}|[^\d]{1,}|[^\W]{1,})$|[\s]
If your password matches the regex above, it is invalid.
If there's no match, your password is valid and contains has at least 8 characters, one upper case letter, one lower case letter and one symbol or special character. And it also contains no spaces, tabs or line breaks.
Breakdown of Regex
.{0,7} - matches if password has between 0 to 7 characters.
[^a-z]{1,} - matches if no lower case is found
[^A-Z]{1,} - matches if no upper case is found
[^\d]{1,} - matches if no number (between [0-9]) is found
[\s] - matches if a white space, tab or line break is found.
With this approach there's no limit or restriction in terms of symbols allowed. If you want to limit to few symbols allowable, just change [^\W] with [^YourSymbols].
(?=.*[0-9])(?=.*[a-z])(?=.*[A-Z])(?=.*[##$%^&+-]).{6}
According to your need this pattern should work just fine. Try this,
^(?=(.*\d){1})(.*\S)(?=.*[a-zA-Z\S])[0-9a-zA-Z\S]{8,}
Just create a string variable, assign the pattern, and create a boolean method which returns true if the pattern is correct, else false.
Sample:
String pattern = "^(?=(.*\d){1})(.*\S)(?=.*[a-zA-Z\S])[0-9a-zA-Z\S]{8,}";
String password_string = "Type the password here"
private boolean isValidPassword(String password_string) {
return password_string.matches(Constants.passwordPattern);
}
Use the following Regex to satisfy the below conditions:
Conditions: 1] Min 1 special character.
2] Min 1 number.
3] Min 8 characters or More
Regex: ^(?=.*\d)(?=.*[#$#!%&*?])[A-Za-z\d#$#!%&*?]{8,}$
Can Test Online: https://regex101.com
Just we can do this by using HTML5.
Use below code in pattern attribute,
pattern="(?=^.{8,}$)((?=.*\d)(?=.*\W+))(?![.\n])(?=.*[A-Z])(?=.*[a-z]).*$"
It will work perfectly.
You can use the below regular expression pattern to check the password whether it matches your expectations or not.
((?=.*\\d)(?=.*[a-z])(?=.*[A-Z])(?=.*[~!##$%^&*()]).{8,20})

How to combine regular expressions in Swift?

I have following individual regular expressions I want to combine them using AND condition , I am using them for validating password
.[A-Z]+. - validate uppercase (one letter uppercase min)
.[0-9]+. - validate number ( one number atleast )
.[a-z]+. - validate lowercase ( one lower case minimum )
.{8,} - validate min character 8
.[^A-Za-z0-9]. - validate special character (atleast one special character )
(.)\1 - validate consecutive characters (no consecutive characters )
Right now I am validating every character separately , but i want to do it in one function only
I tried following way of combining
/^((.)\1)(.[A-Z]+.)(.[a-z]+.)(.[0-9]+.)(.[^A-Za-z0-9].).*$/
Above doesn't have all the expressions but I am trying to show how I have done.
One option is to use a set of positive lookaheads using negated character classes:
^(?=[^A-Z]*[A-Z])(?=[^a-z]*[a-z])(?=[^0-9]*[0-9])(?!.*(.)\1)(?=[a-zA-Z0-9]*[^A-Za-z0-9\s])\S{8,}$
That will match:
^ Start of string
(?=[^A-Z]*[A-Z]) Assert uppercase
(?=[^a-z]*[a-z]) Assert lowercase
(?=[^0-9]*[0-9]) Assert digit
(?!.*(.)\1) Assert no consecutive char
(?=[a-zA-Z0-9]*[^A-Za-z0-9\s]) Assert char other than listed including a whitespace char (assuming that would not be allowed)
\S{8,} Match 8+ times a non whitespace char
$ End of string
Regex demo
Note that \S for the allowed chars is a broad match, you could specify what you would allow to match using a character class.

Using an escaped (magic) character as boundary in a character range in Lua patterns

The Lua manual in section 6.4.1 on Lua Patterns states
A character class is used to represent a set of characters. The
following combinations are allowed in describing a character class:
x: (where x is not one of the magic characters ^$()%.[]*+-?) represents the character x itself.
.: (a dot) represents all characters.
%a: represents all letters.
%c: represents all control characters.
%d: represents all digits.
%g: represents all printable characters except space.
%l: represents all lowercase letters.
%p: represents all punctuation characters.
%s: represents all space characters.
%u: represents all uppercase letters.
%w: represents all alphanumeric characters.
%x: represents all hexadecimal digits.
%x: (where x is any non-alphanumeric character) represents the character x. This is the standard way to escape the magic characters.
Any non-alphanumeric character (including all punctuation characters,
even the non-magical) can be preceded by a % when used to represent
itself in a pattern.
[set]: represents the class which is the union of all characters in set. A range of characters can be specified by separating the end
characters of the range, in ascending order, with a -. All classes
%x described above can also be used as components in set. All other
characters in set represent themselves. For example, [%w_] (or
[_%w]) represents all alphanumeric characters plus the underscore,
[0-7] represents the octal digits, and [0-7%l%-] represents the
octal digits plus the lowercase letters plus the - character.
You can put a closing square bracket in a set by positioning it as the
first character in the set. You can put a hyphen in a set by
positioning it as the first or the last character in the set. (You can
also use an escape for both cases.)
The interaction between ranges and classes is not defined. Therefore, patterns like [%a-z] or [a-%%] have no meaning.
[^set]: represents the complement of set, where set is interpreted
as above.
For all classes represented by single letters (%a, %c, etc.), the
corresponding uppercase letter represents the complement of the class.
For instance, %S represents all non-space characters.
The definitions of letter, space, and other character groups depend on
the current locale. In particular, the class [a-z] may not be
equivalent to %l.
(Highlighting and some formatting added by me)
So, since the "interaction between ranges and classes is not defined.", how do you create a character class set that starts and/or ends with a (magic) character that needs to be escaped?
For example,
[%%-c]
does not define a character class that ranges from % to c and includes all characters in-between but a set that consists only of the three characters %, -, and c.
The interaction between ranges and classes is not defined.
Obviously, this is not a hard and fast rule (of regex character sets in general) but a Lua implementation decision. While using shorthand characters in character sets/ranges work in some (most) regex flavors, it does not in all (like in Python's re module, demo).
However, the second example is misleading:
Therefore, patterns like [%a-z] or [a-%%] have no meaning.
While the first example is fine since %a is a shorthand class (that represents all letters) in a set, [%a-z] is undefined and will return nil if matched against a string.
Escaped range characters in a [set]
In the second example, [a-%%], %% simply defines an escaped % sign and not a shorthand character class. The superficial problem is, the range is defined upsidedown, from high to low (in reference to the US ASCII value of the characters a 61 and % 37), e.g like an erroneous Lua pattern like [f-a]. If the set is defined in reverse order it seems to work: [%%-a] but all it does is matching the three individual characters instead of the range of characters between % and a; credit cyclaminist).
This could be considered a bug and, indeed, means it is not possible to create a range of characters in a [set] if one of the defining range characters need to be escaped.
Possible Solution
Start the character range from the next character that does not need to be escaped - and then add the remaining escaped characters individually, e.g.
[%%&-a]
Sample:
for w in string.gmatch("%&*()-0Aa", "[%%&-a]") do
print(w)
end
This is the answer I have found. Still, maybe somebody else has something better.

Regular Expressions in iOS [duplicate]

I'm creating a regexp for password validation to be used in a Java application as a configuration parameter.
The regexp is:
^.*(?=.{8,})(?=..*[0-9])(?=.*[a-z])(?=.*[A-Z])(?=.*[##$%^&+=]).*$
The password policy is:
At least 8 chars
Contains at least one digit
Contains at least one lower alpha char and one upper alpha char
Contains at least one char within a set of special chars (##%$^ etc.)
Does not contain space, tab, etc.
I’m missing just point 5. I'm not able to have the regexp check for space, tab, carriage return, etc.
Could anyone help me?
Try this:
^(?=.*[0-9])(?=.*[a-z])(?=.*[A-Z])(?=.*[##$%^&+=])(?=\S+$).{8,}$
Explanation:
^ # start-of-string
(?=.*[0-9]) # a digit must occur at least once
(?=.*[a-z]) # a lower case letter must occur at least once
(?=.*[A-Z]) # an upper case letter must occur at least once
(?=.*[##$%^&+=]) # a special character must occur at least once
(?=\S+$) # no whitespace allowed in the entire string
.{8,} # anything, at least eight places though
$ # end-of-string
It's easy to add, modify or remove individual rules, since every rule is an independent "module".
The (?=.*[xyz]) construct eats the entire string (.*) and backtracks to the first occurrence where [xyz] can match. It succeeds if [xyz] is found, it fails otherwise.
The alternative would be using a reluctant qualifier: (?=.*?[xyz]). For a password check, this will hardly make any difference, for much longer strings it could be the more efficient variant.
The most efficient variant (but hardest to read and maintain, therefore the most error-prone) would be (?=[^xyz]*[xyz]), of course. For a regex of this length and for this purpose, I would dis-recommend doing it that way, as it has no real benefits.
simple example using regex
public class passwordvalidation {
public static void main(String[] args) {
String passwd = "aaZZa44#";
String pattern = "(?=.*[0-9])(?=.*[a-z])(?=.*[A-Z])(?=.*[##$%^&+=])(?=\\S+$).{8,}";
System.out.println(passwd.matches(pattern));
}
}
Explanations:
(?=.*[0-9]) a digit must occur at least once
(?=.*[a-z]) a lower case letter must occur at least once
(?=.*[A-Z]) an upper case letter must occur at least once
(?=.*[##$%^&+=]) a special character must occur at least once
(?=\\S+$) no whitespace allowed in the entire string
.{8,} at least 8 characters
All the previously given answers use the same (correct) technique to use a separate lookahead for each requirement. But they contain a couple of inefficiencies and a potentially massive bug, depending on the back end that will actually use the password.
I'll start with the regex from the accepted answer:
^(?=.*[0-9])(?=.*[a-z])(?=.*[A-Z])(?=.*[##$%^&+=])(?=\S+$).{8,}$
First of all, since Java supports \A and \z I prefer to use those to make sure the entire string is validated, independently of Pattern.MULTILINE. This doesn't affect performance, but avoids mistakes when regexes are recycled.
\A(?=.*[0-9])(?=.*[a-z])(?=.*[A-Z])(?=.*[##$%^&+=])(?=\S+$).{8,}\z
Checking that the password does not contain whitespace and checking its minimum length can be done in a single pass by using the all at once by putting variable quantifier {8,} on the shorthand \S that limits the allowed characters:
\A(?=.*[0-9])(?=.*[a-z])(?=.*[A-Z])(?=.*[##$%^&+=])\S{8,}\z
If the provided password does contain a space, all the checks will be done, only to have the final check fail on the space. This can be avoided by replacing all the dots with \S:
\A(?=\S*[0-9])(?=\S*[a-z])(?=\S*[A-Z])(?=\S*[##$%^&+=])\S{8,}\z
The dot should only be used if you really want to allow any character. Otherwise, use a (negated) character class to limit your regex to only those characters that are really permitted. Though it makes little difference in this case, not using the dot when something else is more appropriate is a very good habit. I see far too many cases of catastrophic backtracking because the developer was too lazy to use something more appropriate than the dot.
Since there's a good chance the initial tests will find an appropriate character in the first half of the password, a lazy quantifier can be more efficient:
\A(?=\S*?[0-9])(?=\S*?[a-z])(?=\S*?[A-Z])(?=\S*?[##$%^&+=])\S{8,}\z
But now for the really important issue: none of the answers mentions the fact that the original question seems to be written by somebody who thinks in ASCII. But in Java strings are Unicode. Are non-ASCII characters allowed in passwords? If they are, are only ASCII spaces disallowed, or should all Unicode whitespace be excluded.
By default \s matches only ASCII whitespace, so its inverse \S matches all Unicode characters (whitespace or not) and all non-whitespace ASCII characters. If Unicode characters are allowed but Unicode spaces are not, the UNICODE_CHARACTER_CLASS flag can be specified to make \S exclude Unicode whitespace. If Unicode characters are not allowed, then [\x21-\x7E] can be used instead of \S to match all ASCII characters that are not a space or a control character.
Which brings us to the next potential issue: do we want to allow control characters? The first step in writing a proper regex is to exactly specify what you want to match and what you don't. The only 100% technically correct answer is that the password specification in the question is ambiguous because it does not state whether certain ranges of characters like control characters or non-ASCII characters are permitted or not.
You should not use overly complex Regex (if you can avoid them) because they are
hard to read (at least for everyone but yourself)
hard to extend
hard to debug
Although there might be a small performance overhead in using many small regular expressions, the points above outweight it easily.
I would implement like this:
bool matchesPolicy(pwd) {
if (pwd.length < 8) return false;
if (not pwd =~ /[0-9]/) return false;
if (not pwd =~ /[a-z]/) return false;
if (not pwd =~ /[A-Z]/) return false;
if (not pwd =~ /[%#$^]/) return false;
if (pwd =~ /\s/) return false;
return true;
}
Thanks for all answers, based on all them but extending sphecial characters:
#SuppressWarnings({"regexp", "RegExpUnexpectedAnchor", "RegExpRedundantEscape"})
String PASSWORD_SPECIAL_CHARS = "##$%^`<>&+=\"!ºª·#~%&'¿¡€,:;*/+-.=_\\[\\]\\(\\)\\|\\_\\?\\\\";
int PASSWORD_MIN_SIZE = 8;
String PASSWORD_REGEXP = "^(?=.*[0-9])(?=.*[a-z])(?=.*[A-Z])(?=.*[" + PASSWORD_SPECIAL_CHARS + "])(?=\\S+$).{"+PASSWORD_MIN_SIZE+",}$";
Unit tested:
Password Requirement :
Password should be at least eight (8) characters in length where the system can support it.
Passwords must include characters from at least two (2) of these groupings: alpha, numeric, and special characters.
^.*(?=.{8,})(?=.*\d)(?=.*[a-zA-Z])|(?=.{8,})(?=.*\d)(?=.*[!##$%^&])|(?=.{8,})(?=.*[a-zA-Z])(?=.*[!##$%^&]).*$
I tested it and it works
For anyone interested in minimum requirements for each type of character, I would suggest making the following extension over Tomalak's accepted answer:
^(?=(.*[0-9]){%d,})(?=(.*[a-z]){%d,})(?=(.*[A-Z]){%d,})(?=(.*[^0-9a-zA-Z]){%d,})(?=\S+$).{%d,}$
Notice that this is a formatting string and not the final regex pattern. Just substitute %d with the minimum required occurrences for: digits, lowercase, uppercase, non-digit/character, and entire password (respectively). Maximum occurrences are unlikely (unless you want a max of 0, effectively rejecting any such characters) but those could be easily added as well. Notice the extra grouping around each type so that the min/max constraints allow for non-consecutive matches. This worked wonders for a system where we could centrally configure how many of each type of character we required and then have the website as well as two different mobile platforms fetch that information in order to construct the regex pattern based on the above formatting string.
This one checks for every special character :
^(?=.*[0-9])(?=.*[a-z])(?=.*[A-Z])(?=\S+$).*[A-Za-z0-9].{8,}$
Java Method ready for you, with parameters
Just copy and paste and set your desired parameters.
If you don't want a module, just comment it or add an "if" as done by me for special char
//______________________________________________________________________________
/**
* Validation Password */
//______________________________________________________________________________
private static boolean validation_Password(final String PASSWORD_Arg) {
boolean result = false;
try {
if (PASSWORD_Arg!=null) {
//_________________________
//Parameteres
final String MIN_LENGHT="8";
final String MAX_LENGHT="20";
final boolean SPECIAL_CHAR_NEEDED=true;
//_________________________
//Modules
final String ONE_DIGIT = "(?=.*[0-9])"; //(?=.*[0-9]) a digit must occur at least once
final String LOWER_CASE = "(?=.*[a-z])"; //(?=.*[a-z]) a lower case letter must occur at least once
final String UPPER_CASE = "(?=.*[A-Z])"; //(?=.*[A-Z]) an upper case letter must occur at least once
final String NO_SPACE = "(?=\\S+$)"; //(?=\\S+$) no whitespace allowed in the entire string
//final String MIN_CHAR = ".{" + MIN_LENGHT + ",}"; //.{8,} at least 8 characters
final String MIN_MAX_CHAR = ".{" + MIN_LENGHT + "," + MAX_LENGHT + "}"; //.{5,10} represents minimum of 5 characters and maximum of 10 characters
final String SPECIAL_CHAR;
if (SPECIAL_CHAR_NEEDED==true) SPECIAL_CHAR= "(?=.*[##$%^&+=])"; //(?=.*[##$%^&+=]) a special character must occur at least once
else SPECIAL_CHAR="";
//_________________________
//Pattern
//String pattern = "(?=.*[0-9])(?=.*[a-z])(?=.*[A-Z])(?=.*[##$%^&+=])(?=\\S+$).{8,}";
final String PATTERN = ONE_DIGIT + LOWER_CASE + UPPER_CASE + SPECIAL_CHAR + NO_SPACE + MIN_MAX_CHAR;
//_________________________
result = PASSWORD_Arg.matches(PATTERN);
//_________________________
}
} catch (Exception ex) {
result=false;
}
return result;
}
Also You Can Do like This.
public boolean isPasswordValid(String password) {
String regExpn =
"^(?=.*[0-9])(?=.*[a-z])(?=.*[A-Z])(?=.*[##$%^&+=])(?=\\S+$).{8,}$";
CharSequence inputStr = password;
Pattern pattern = Pattern.compile(regExpn,Pattern.CASE_INSENSITIVE);
Matcher matcher = pattern.matcher(inputStr);
if(matcher.matches())
return true;
else
return false;
}
Use Passay library which is powerful api.
I think this can do it also (as a simpler mode):
^(?=.*\d)(?=.*[a-z])(?=.*[A-Z])(?=.*[##$%^&+=])[^\s]{8,}$
[Regex Demo]
easy one
("^ (?=.* [0-9]) (?=.* [a-z]) (?=.* [A-Z]) (?=.* [\\W_])[\\S]{8,10}$")
(?= anything ) ->means positive looks forward in all input string and make sure for this condition is written .sample(?=.*[0-9])-> means ensure one digit number is written in the all string.if not written return false
.
(?! anything ) ->(vise versa) means negative looks forward if condition is written return false.
close meaning ^(condition)(condition)(condition)(condition)[\S]{8,10}$
String s=pwd;
int n=0;
for(int i=0;i<s.length();i++)
{
if((Character.isDigit(s.charAt(i))))
{
n=5;
break;
}
else
{
}
}
for(int i=0;i<s.length();i++)
{
if((Character.isLetter(s.charAt(i))))
{
n+=5;
break;
}
else
{
}
}
if(n==10)
{
out.print("Password format correct <b>Accepted</b><br>");
}
else
{
out.print("Password must be alphanumeric <b>Declined</b><br>");
}
Explanation:
First set the password as a string and create integer set o.
Then check the each and every char by for loop.
If it finds number in the string then the n add 5. Then jump to the
next for loop. Character.isDigit(s.charAt(i))
This loop check any alphabets placed in the string. If its find then
add one more 5 in n. Character.isLetter(s.charAt(i))
Now check the integer n by the way of if condition. If n=10 is true
given string is alphanumeric else its not.
Sample code block for strong password:
(?=.*[0-9])(?=.*[a-z])(?=.*[A-Z])(?=.*[^a-zA-Z0-9])(?=\\S+$).{6,18}
at least 6 digits
up to 18 digits
one number
one lowercase
one uppercase
can contain all special characters
RegEx is -
^(?:(?=.*\d)(?=.*[a-z])(?=.*[A-Z])(?=.*[##$%^&+=]).*)[^\s]{8,}$
at least 8 digits {8,}
at least one number (?=.*\d)
at least one lowercase (?=.*[a-z])
at least one uppercase (?=.*[A-Z])
at least one special character (?=.*[##$%^&+=])
No space [^\s]
A more general answer which accepts all the special characters including _ would be slightly different:
^(?=.*[0-9])(?=.*[a-z])(?=.*[A-Z])(?=.*[\W|\_])(?=\S+$).{8,}$
The difference (?=.*[\W|\_]) translates to "at least one of all the special characters including the underscore".

Write a Lex rule to parse Integer and Float

I am writing a parse for a script language.
I need to recognize strings, integers and floats.
I successfully recognize strings with the rule:
[a-zA-Z0-9_]+ {return STRING;}
But I have problem recognizing Integers and Floats. These are the (wrong) rules I wrote:
["+"|"-"][1-9]{DIGIT}* { return INTEGER;}
["+"|"-"]["0." | [1-9]{DIGIT}*"."]{DIGIT}+ {return FLOAT;}
How can I fix them?
Furthermore, since a "abc123" is a valid string, how can I make sure that it is recognized as a string and not as the concatenation of a string ("abc") and an Integer ("123") ?
First problem: There's a difference between (...) and [...]. Your regular expressions don't do what you think they do because you're using the wrong punctuation.
Beyond that:
No numeric rule recognizes 0.
Both numeric rules require an explicit sign.
Your STRING rule recognizes integers.
So, to start:
[...] encloses a set of individual characters or character ranges. It matches a single character which is a member of the set.
(...) encloses a regular expression. The parentheses are used for grouping, as in mathematics.
"..." encloses a sequence of individual characters, and matches exactly those characters.
With that in mind, let's look at
["+"|"-"][1-9]{DIGIT}*
The first bracket expression ["+"|"-"] is a set of individual characters or ranges. In this case, the set contains: ", +, " (again, which has no effect because a set contains zero or one instances of each member), |, and the range "-", which is a range whose endpoints are the same character, and consequently only includes that character, ", which is already in the set. In short, that was equivalent to ["+|]. It will match one of those three characters. It requires one of those three characters, in fact.
The second bracket expression [1-9] matches one character in the range 1-9, so it probably does what you expected. Again, it matches exactly one character.
Finally, {DIGIT} matches the expansion of the name DIGIT. I'll assume that you have the definition:
DIGIT [0-9]
somewhere in your definitions section. (In passing, I note that you could have just used the character class [:digit:], which would have been unambiguous, and you would not have needed to define it.) It's followed by a *, which means that it will match zero or more repetitions of the {DIGIT} definition.
Now, an example of a string which matches that pattern:
|42
And some examples of strings which don't match that pattern:
-7 # The pattern must start with |, + or "
42 # Again, the pattern must start with |, + or "
+0 # The character following the + must be in the range [0-9]
Similarly, your float pattern, once the [...] expressions are simplified, becomes (writing out the individual pieces one per line, to make it more obvious):
["+|] # i.e. the set " + |
["0.|[1-9] # i.e. the set " 0 | [ 1 2 3 4 5 6 7 8 9
{DIGIT}* # Any number of digits
"." # A single period
] # A single ]
{DIGIT}+ # one or more digits
So here's a possible match:
"..]3
I'll skip over writing out the solution because I think you'll benefit more from doing it yourself.
Now, the other issues:
Some rule should match 0. If you don't want to allow leading zeros, you'll need to just a it as a separate rule.
Use the optional operator (?) to indicate that the preceding object is optional. eg. "foo"? matches either the three characters f, o, o (in order) or matches the empty string. You can use that to make the sign optional.
The problem is not the matching of abc123, as in your question. (F)lex always gives you the longest possible match, and the only rule which could match the starting character a is the string rule, so it will allow the string rule to continue as long as it can. It will always match all of abc123. However, it will also match 123, which you would probably prefer to be matched by your numeric rule. Here, the other (f)lex matching criterion comes into play: when there are two or more rules which could match exactly the same string, and none of the rules can match a longer string, (f)lex chooses the first rule in the file. So if you want to give numbers priority over strings, you have to put the number rule earlier in your (f)lex file than the string rule.
I hope that gives you some ideas about how to fix things.

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