Regular expression for Password match in PHP and Javascript

Question

I have built a regular expression for password policy match, but it is not working as expected

/((?=.\d)(?=.[a-z])(?=.[A-Z])(?=.[@#\$%!])(?!(.)*\1{2,}).{6,20})/

Password must satisfy below rules-> must have 1 digit-> must have 1 upper case letter-> must have 1 lower case letter-> must have 1 special character from given list-> minimum 6 character long-> maximum 20 character long-> Not more than 2 identical characters`

So it matches

aDm!n1, Adw1n@

but it must not match below aaaD!n1, teSt@111

I have searched for this regular expression and found "(?!(.)*\1{2,})" is not working properly

I am not getting why it is not working even though it has lookahead negative assertion.

Thanks in advance

"Not more than 2 identical characters" --- this is a stupid rule. "maximum 20 character long" --- this is another stupid rule. — zerkms
– zerkms, Commented Nov 28, 2014 at 6:39
So you intentionally added meaningful (and insecure) requirements? That's weird. — zerkms
– zerkms, Commented Nov 28, 2014 at 6:42
How it is insecure? there is no maximum character limit actually. It is to stop user entering passwords like [email protected] — Gohel Kiran
– Gohel Kiran, Commented Nov 28, 2014 at 6:44
"there is no maximum character limit actually" --- .{6,20} --- 1. there is 2. It reduces the space of possible passwords significantly. Do you seriously think this is ziMsp8DyQk3MXK3CDNOocsDOHHHH an insecure password? — zerkms
– zerkms, Commented Nov 28, 2014 at 6:44

Avinash Raj · Accepted Answer · 2014-11-28 06:50:38Z

1

You must need to provide start and end anchors.

^(?=.*\d)(?=.*[a-z])(?=.*[A-Z])(?=.*[@#\$%\!])(?!.*(.).*\1.*\1).{6,20}$

DEMO

To match the strings which isn't contain more than two duplicate characters, you need to use a negative lookahead like (?!.*(.).*\1.*\1) which asserts that the string we are going to match wouldn't contain not more than two duplicate characters.

(?!) Negative lookahead which checks if there isn't
.* Any character zero or more times.
(.) A single character was captured.
.* Any character zero or more times.
\1 Reference to the group index 1. That is, it refers to the character which are already captured by group 1.
.* Any character zero or more times.
\1 Back-referencing to the character which was present inside the group index 1.

Code:

> var re = /^(?=.*\d)(?=.*[a-z])(?=.*[A-Z])(?=.*[@#\$%\!])(?!.*(.).*\1.*\1).{6,20}$/;
undefined
> re.test('aDm!n1')
true
> re.test('Adw1n@')
true
> re.test('tetSt@11')
false

edited Nov 28, 2014 at 6:50

answered Nov 28, 2014 at 6:19

Avinash Raj

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5 Comments

Gohel Kiran Over a year ago

teSt@111 this is invalid as it contains 111 thrice. But it is still validated

Avinash Raj Over a year ago

@user1531437 now check my regex.

Gohel Kiran Over a year ago

Still it is not working, gives "Your pattern does not match the subject string."

Avinash Raj Over a year ago

could you provide the subject string? If you don't want to allow the strings like ftftT (ie, there t's upper or lower case) then use a case insensitive modifier.

Gohel Kiran Over a year ago

It must not allow same continuous character more than twice abababab Allowed aaaabbcc Not allowed It has nothing to do with case, any character must be checcked

vks · Accepted Answer · 2014-11-28 06:43:08Z

0

(?:(?=.*\d)(?=.*[a-z])(?=.*[A-Z])(?=.*[@#\$%!])(?!.*(.).*\1.*\1).{6,20})

You need this.See demo.

http://regex101.com/r/hQ9xT1/22

Your regex was failing cos

(?!(.)*\1{2,}) will not work as it find consecutive repeated characters and not any character which is three times or more.So use (?!.*(.).*\1.*\1).

var re = /^(?:(?=.*\d)(?=.*[a-z])(?=.*[A-Z])(?=.*[@#\$%!])(?!.*(.).*\1.*\1).{6,20})$/gm;
var str = 'aaaD!n1\nteSt@111\naDm!n1\nAdw1n@';
var m;

while ((m = re.exec(str)) != null) {
if (m.index === re.lastIndex) {
re.lastIndex++;
}
// View your result using the m-variable.
// eg m[0] etc.
}

edited Nov 28, 2014 at 6:43

answered Nov 28, 2014 at 6:14

vks

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