username is not a part of password and vice versa javascript or jquery

Question

Password should not contain username or parts of the user's full name that exceed two consecutive characters

If username= 1Abcd234 password= aBc15 then it should return error like password could not contain any parts of username.

Because "abc1" is present in both password and username.

And what have you tried to make this, please note that StackOverflow is not a coding service — Carsten Løvbo Andersen, Oct 13 '17 at 05:05
I use this code function check(user,pswd){ use= user.toLowerCase(); pwd= pswd.toLowerCase(); var value= (pwd.match(/[a-z]+/ig).filter(a=> a.length > 1 && use.includes(a)).length > 0); return value; } if(check("abcdefgh","abc1")==true){ alert('true'); } else { alert('false'); } But in this if I use username like abcsfgh then it return false. — Nishu Kashyap, Oct 13 '17 at 05:15
please check this answer: https://stackoverflow.com/a/36566052/5060513 it check between 2 strings similarity in ratio 0 to 1 — Taufik Nur Rahmanda, Oct 13 '17 at 05:28

Derek Brown · Answer 1 · 2017-10-13T05:43:19.480

0

I think you could use something like this:

function FindIntersectionFromStart(a,b){
    for(var i=a.length;i>0;i--){
        d = a.substring(0,i);
        j = b.indexOf(d);
        if (j>=0){
            return ({position:j,length:i});
        }
    }

    return null;
}

function FindIntersection(a,b){
    var bestResult = null;
    for(var i=0;i<a.length-1;i++){
        var result = FindIntersectionFromStart(a.substring(i),b);
        if (result){
            if (!bestResult){
                bestResult = result;
            } else {
                if (result.length>bestResult.length){
                    bestResult = result;
                }
            }
        }
        if(bestResult && bestResult.length>=a.length-i)
            break;
    }
    return bestResult;
}

var username = "myUsername";
var password = "myuse";

result = FindIntersection(username.toLowerCase(), password.toLowerCase());

if(result.length > 2){
  console.log("Invalid Password!");
} else {
  console.log("Valid Password!")
}

edited Oct 13 '17 at 05:43

answered Oct 13 '17 at 05:23

Derek Brown

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Sorry but if I use this var username = "myUsername"; var password = "myuse"; It will return valid password . – Nishu Kashyap Oct 13 '17 at 05:32
@NishuKashyap this occurs because the cases are different. By adding `.toLowerCase()` your example works. – Derek Brown Oct 13 '17 at 05:44
Sorry but you don't get my point. If myUsername and myuse both have same string i.e. myuse then how can return valid password. – Nishu Kashyap Oct 13 '17 at 05:56
@NishuKashyap I updated the code to meet your example. Try it again. – Derek Brown Oct 13 '17 at 05:57
Thank you so much. Now It works fine for me. @Derek Brown – Nishu Kashyap Oct 13 '17 at 06:08
@Kashyap this code currently does this- it returns valid iff there is no common substring with length greater than 2. – Derek Brown Oct 13 '17 at 07:46
Yes I also get this. If you have any solution for this also then please let me know. Thanks again @Derek Brown – Nishu Kashyap Oct 13 '17 at 09:37

Mohit Bhardwaj · Answer 2 · 2017-10-13T06:00:18.070

Here's a simple approach (may not be the best though):

First get all combinations of consecutive characters from username that should not be found in password. This would depend on the number of consecutive characters that your application considers invalid if matched.
Now, run a simple loop and check the presence of any of the invalid combinations using String.indexOf method.
Make sure that you convert both - username and password to same case for simplicity.

$("#check").on("click", checkValidity );

function checkValidity(){
  var numConsecutiveChars = 2;
  var username = $("#username").val().trim().toLowerCase();
  var password = $("#password").val().trim().toLowerCase();
  
  // first find all combinations that should not be found in password
  var invalidCombinations = [];
  for( var i = 0; i < username.length - numConsecutiveChars; i++ ){
    var curCombination = username[i] + username[i+1];
    invalidCombinations.push( curCombination );
  }//for
  
  // now check all invalidCombinations
  var invalid = false;
  for( var i = 0; i < invalidCombinations.length; i++ ){
    var curCombination = invalidCombinations[i];
    if( password.indexOf( curCombination ) !== -1 ){
      invalid = true;
      break;
    }
  }//for()
  
  if( invalid ){
    alert("Invalid password");
  }else{
    alert("Valid password");
  }
}//checkValidity()

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<input type="text" id="username" value="user" />
<input type="text" id="password" value="pass" />
<button id="check">Check validity</button>

Yes this code also works. Thank you so much. @mohit bhardwaj — Nishu Kashyap, Oct 13 '17 at 06:42

username is not a part of password and vice versa javascript or jquery

2 Answers2