Password Validation with Sequential Letters and Nu

Question

To have customer accounts more secure, a well crafted password is good practice. This is my Regular Expression string for password validation.

/^(?=.*[0-9])(?!.*?\d{3})(?=.*[a-zA-Z])(?!.*?[a-zA-Z]{3})(?=.*[~!@#$%^&*()+-?])([a-zA-Z0-9~!@#$%^&*()+-?]{8,})$/

Represents:

• 8 or more characters.
• Uppercase letter A-Z
• Lowercase letters a-z
• Special characters ~!@#$%^&*()+-?
• What is this Regular Expression function for this?: Must not contain up to 3 sequential letters and/or numbers.

Having numbers and/or letters in order 3 or more sequential is not OK.

Example:

Not OK = efg123!$, abcd567%, xyz789^&, #hijk23456
OK = ryiiu562@, erty745#, gjnfl45566^

Thank you

Answer 1

You can have a function similar to the below by looping the characters and using a charCodeAt string method as below.

Note: This is for the question raised in below link as well.

string validation for 3 or more consecutive sequential alphanumeric characters in javascript

function validate() {
  var pwd = document.getElementById('password').value;
  var isValid = checkPassword(pwd);
  var elm = document.getElementById('result');
  elm.innerHTML = isValid ? 'Valid' : 'Invalid';
  elm.style.color = isValid ? 'green' : 'red';
}

function checkPassword(s) {
    
    if(s) {
       var test = (x) => !isNaN(x);
       var check = (x, y, i) => x + i === y;
    
       for(var i = 0; i < s.length - 2; i++) {
         if(test(s[i])) {
            if(test(s[i + 1]) && test(s[i + 2])) {
              if(check(s[i], s[i + 1], 1) &&
                check(s[i], s[i + 2], 2)) {
                return false;
              }
            }
         } else if(!test(s[i + 1]) && !test(s[i + 2])) {
            if(check(s.charCodeAt(i), s.charCodeAt(i + 1), 1) &&
                check(s.charCodeAt(i), s.charCodeAt(i + 2), 2)) {
                return false;
              }
         }
       }
      
    }
    
    return true;
}

document.getElementById('buttonToValidate').click();

<input type="text" id="password" value="efg123!$" /> 
<input type="button" id="buttonToValidate" value="Check" onclick="validate()" />
<span id="result" />

Answer 2

There's no way using RegEx that I know of, but here is a naive functional approach.

First, loop through the string and compare each character against the next two characters by adding +1 and +2 to the current index and comparing appropriately.

Second, loop through the string again and compare checks the next two characters against the current character to see if they are sequential.

If both loops fail to find sequential characters, the function returns true, otherwise it returns false.

The first four return false (fail), while the last three return true (pass).

function test(s) {
    // Check for sequential numerical characters
    for(var i in s) 
        if (+s[+i+1] == +s[i]+1 && 
            +s[+i+2] == +s[i]+2) return false;
    // Check for sequential alphabetical characters
    for(var i in s) 
        if (String.fromCharCode(s.charCodeAt(i)+1) == s[+i+1] && 
            String.fromCharCode(s.charCodeAt(i)+2) == s[+i+2]) return false;
    return true;
}

// For demo purposes only
var tests = [
    'efg123!$',
    'abcd567%',
    'xyz789^&',
    '#hijk23456',
    'ryiiu562@',
    'erty745#',
    'gjnfl45566^'
], sep = '\t\u2192 ', out = ['Fail','Pass'], eol = '<br>';
document.write('<pre>');
for(var i in tests) document.write(tests[i] + sep + out[+test(tests[i])] + eol);
document.write('</pre>');