Regular Expression for Password Strength Validation

Question

I have written a regular expression which could potentially be used for password strength validation:

^(?:([A-Z])*([a-z])*(\\d)*(\\W)*){8,12}$

Answer 1

You can do it by separating each group out and match them one by one, like this:

var level = 0;
var input = '';//user input goes here
switch(true){
    case /^(?:([A-Z])*){8,12}$/.test(input):
    level = 1;
    break;

    case /^(?:([A-Z])*([a-z])*){8,12}$/.test(input):
    level = 2;
    break;

    case /^(?:([A-Z])*([a-z])*(\d)*){8,12}$/.test(input):
    level = 3;
    break;

    case /^(?:([A-Z])*([a-z])*(\d)*(\W)*){8,12}$/.test(input):
    level = 4;
    break;
}

The level variable goes from 1 (the weakest) to 4 (the strongest).

Answer 2

Check out this: http://jsfiddle.net/43tu58jf/

function isSimpleStrongLevel(password){
    var stringsOptions = ['123456', '123abc', 'abc123', '654321', '012345', 'password', '123pass', 'pass123', '123456abc'];
    return stringsOptions.indexOf(password) != -1;
}
function getStrongLevel(password) {
    var level = 0;
    level += password.length > 6 ? 1 : 0;
    level += /[!@#$%^&*?_~]{2,}/.test(password) ? 1 : 0;
    level += /[a-z]{2,}/.test(password) ? 1 : 0;
    level += /[A-Z]{2,}/.test(password) ? 1 : 0;
    level += /[0-9]{2,}/.test(password) ? 1 : 0;
    return level;
}

var password = '123456';
var isSimple = isSimpleStrongLevel(password);
var level = getStrongLevel(password);
console.log(isSimple, level);

Answer 3

Try it ==>

var PasswordStrength = function () {
    var p_worker = {
        checkMark: function ( msg, mark ) {
            let strength = "Required!!!", status = true;
            switch ( mark ) {
                case 1: strength = '<b style="color:rgb(200, 200, 200)"> Password strength: Weak...</b>'; break;
                case 2: strength = '<b style="color:rgb(200, 200, 200)"> Password strength: Semi-weak...</b>'; break;
                case 3: strength = '<b style="color:green"> Password strength: Medium...</b>'; break;
                case 4: strength = '<b style="color:green"> Password strength: Strong...</b>'; break;
                default: status = false; break;
            }
            return { status: status/*[is valid or not]*/, cur_strength: strength/**[strength msg]*/, req_msg: msg/**[required msg]*/, mark: mark/**[strength mark]*/ };
        },
        setting: {
            n: { rgx: /[0-9]/, msg: '1 Numeric character' },
            c: { rgx: /[A-Z]/, msg: '1 Alphabet character' },
            s: { rgx: /[a-z]/, msg: '1 Small character' },
            sp: { rgx: /[@#$\.%^&+=]/, msg: '1 Special character' },
        }
    };
    return {
        check: function ( value ) {
            let msg = "", mark = 0, c = 0;
            for ( let i in p_worker.setting ) {
                if ( !p_worker.setting[i]['rgx'].test( value ) ) {
                    if ( c === 3 ) {
                        msg += '<d style="color:rgba(219, 177, 177, 0.96)">[*] ' + p_worker.setting[i]['msg'] + '</d>';
                        c++; continue;
                    }
                    msg += '<d style="color:rgba(219, 177, 177, 0.96)">[*] ' + p_worker.setting[i]['msg'] + ',</d></br>';
                    c++; continue;
                }
                if ( c === 3 ) {
                    msg += '<img src="/image/accept.png" /> <d style="color:green">' + p_worker.setting[i]['msg'] + '</d>';
                    mark++; c++; continue;
                }
                msg += '<img src="/image/accept.png" /> <d style="color:green">' + p_worker.setting[i]['msg'] + ',</d></br>';
                mark++; c++;
            }
            return p_worker.checkMark( msg, mark );
        }
    }
}();

Use ==>

PasswordStrength.check( "1234$#" );

Answer 4

I think you'll have to check each group independently. Pseudo-code:

bool[] array = {};
array[0] = pwd.match(/[A-Z]/);
array[1] = pwd.match(/[a-z]/);
array[2] = pwd.match(/\d/);
array[3] = pwd.match(/[!_.-]/);

int sum = 0;
for (int i=0; i<array.length; i++) {
    sum += array[i] ? 1 : 0;
}

switch (sum) {
    case 0: print("weird..."); break;
    case 1: print("weak"); break;
    case 2: print("ok"); break;
    case 3: print("strong"); break;
    case 4: print("awesome"); break;
    default: print("weird..."); break;
}

Answer 5

Here was my final solution based on sp00m's answer:

function testPassword(pwString) {
    var strength = 0;

    strength += /[A-Z]+/.test(pwString) ? 1 : 0;
    strength += /[a-z]+/.test(pwString) ? 1 : 0;
    strength += /[0-9]+/.test(pwString) ? 1 : 0;
    strength += /[\W]+/.test(pwString) ? 1 : 0;

    switch(strength) {
        case 3:
            // its's medium!
            break;
        case 4:
            // it's strong!
            break;
        default:
            // it's weak!
            break;
    }
}

I've added this purely for reference, however have accepted sp00m's answer since it was their answer that let me to this solution.

Answer 6

my example

JS/jQuery

$( "#password" ).change(function ()
{
    strongPassword();
});

/**
 * @author websky
 */
function strongPassword() {
    var p = document.getElementById('password').value;
    var label1 = 0; 
    var label2 = 0; 
    var label3 = 0; 
    var label4 = 0; 
    var label5 = 0;
    if (p.length > 6) {//min length
        label1 = 1;
    }
    if (/[!@#$%^&*?_~]{2,}/.test(p)) {//min 2 special characters
        label2 = 1;
    }
    if (/[a-z]{2,}/.test(p)) {//min 2 a-z
        label3 = 1;
    }
    if (/[A-Z]{2,}/.test(p)) {//min 2 A-Z
        label4 = 1;
    }
    if (/[0-9]{2,}/.test(p)) {//min 2 0-9
        label5 = 1; 
    }

    var strong_password = label1 + label2 + label3 + label4 + label5;

    if(strong_password > 0){
        //Here your action
        //
        //var progressbar = strong_password * 20; 
        //$( "#progressbar" ).progressbar({value: progressbar}); <== I use jQuery progessbar
    } 
}