Regex with no consecutive letters and numbers

Question

I have a requirement to make a regex that verifies the following points for an alphanumeric string in Swift and I have no clue on how to achieve them:

• No more than 4 consecutive alphabetic or numeric characters

  (eg 12345a or 1abcde would be wrong but 123a45 or abcd1e would be correct)

• No more than 4 of the same alphanumeric characters

  (eg 111a11 or aaaa1a would be wrong)

• Exactly 6 characters

Help would be greatly appreciated, thanks.

Answer 1

As pointed out by others, your second requirement is hard to accomplish with regular expressions alone while the first and third can be done with:

^
(?!\D*\d{5,}\D*)
(?![^A-Za-z]*[A-Za-z]{5,}[^A-Za-z]*)
.{6}
$

See a demo on regex101.com .

Answer 2

To do the whole job, in each simple step: Note: This is a javascript solution but you could translate to your language of choice following the same logic:

var str = "ss4sss";

fnTest = function(str){
  var isValid = true;


  if(/\d{5,}/.test(str)){// no more than 4 consecutive numeric chars
    return false;
  }else if(/[a-z]{5,}/i.test(str)){// no more than 4 consecutive alpha chars
    return false;
  }else if(str.length !== 6){// not 6 characters
    return false;    
  }else{ // any character used more than 4 times
    var reg = /([\w\W])/g;
    var arrM ;
    var obj = {};
    while(arrM = reg.exec(str)){
      var letter = arrM[1];
      if(!obj[letter]){
        obj[letter] = 0;
      }
      obj[letter]++;

      if(obj[letter] > 4){    
        isValid = false;
        break;
      }
    }
  }
  return isValid;
}
fnTest(str);