Selecting all characters in string between nth number and nth number Regex

Question

I am trying to use regex in a jQuery function to select and mask all characters in a string with an'x' except the first 4 and last 4 characters. The string can be any length. I can successfully mask the last 4 digits and first 4 digits separately but I don't really understand regex well enough to select the nth character in a string to the nth character and mask them. If anybody can help it would be very grateful - I have spent many hours trawling around forums and trying to write my own regex but to no avail.

Thanks

My current function looks like this:

<input type="text" class="read-only-mask" title="" value="1233434343434456789012" name="" id=""readonly />
<script>
$(document).ready(function() {
    $('.read-only-mask').val(function(_,val) {
         return val.replace(/.(?=.{4})/g, 'x');     
    });
});         
</script>

This would show 1233434343434456789012 as xxxxxxxxxxxxxxxxxx9012

I need it to show as 1233xxxxxxxxxxxxxx9012 but the string could be any length so 123343434343 would need to show as 1233****4343 etc

why use regex or any other thing for this simple loop? for(var i = 4; i < 10; i++) { string[i] = maskChar; } — Batman
– Batman, Commented Jun 8, 2016 at 15:25
@RichardBarker, I really think these types of problems can be summed up by a friend of mine. He once said that we as engineers build solutions that challenge our intellect. I am very guilty of this. Thank you for pointing out the obvious. Add it as an answer and I'll vote for you. — Mike Perrenoud
– Mike Perrenoud, Commented Jun 8, 2016 at 15:27
@Jamiec, while his exact algorithm may not have worked, the simplicity by which it was produced is the importance. It may have to copy the data to another string, but that's no big deal. — Mike Perrenoud
– Mike Perrenoud, Commented Jun 8, 2016 at 15:28

Batman · Accepted Answer · 2016-06-08 15:32:25Z

2

you are far better off using a simpler approach. Save your self some time and headache and use the KISS method.

var maxMaskLength = 10;
var minMaskLength = 4;
$(document).ready(function() {
    $('.read-only-mask').val(function(_, val) {
  var valSplit = val.split("");
  for (i = minMaskLength; i < maxMaskLength; i++)
    valSplit[i] = 'x';
  return valSplit.join("");
   });
 });

answered Jun 8, 2016 at 15:32

Batman

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Comments

Redu · Accepted Answer · 2016-06-08 15:38:05Z

2

I would prefer to do it like this

var str = "1233434343434456789012",
    res = Array.prototype.map.call(str, (e,i) => i > 4 && i < 9 ? "X" : e).join("");
console.log(res);

    var str = "1233434343434456789012",
        res = Array.prototype.map.call(str, (e,i,a) => i < a.length-4 ? "X" : e).join("");
    console.log(res);

whichever best fits your application.

edited Jun 8, 2016 at 15:38

answered Jun 8, 2016 at 15:32

Redu

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Comments

Praveen Kumar Purushothaman · Accepted Answer · 2016-06-08 15:28:34Z

1

I would use this way:

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<input type="text" class="read-only-mask" title="" value="1233434343434456789012" name="" id=""readonly />
<script>
  $(document).ready(function() {
    $('.read-only-mask').val(function(_, val) {
      val = val.split("");
      for (i = 4; i < val.length; i++)
        val[i] = 'x';
      return val.join("");
    });
  });         
</script>

For the above input, it shows 1233xxxxxxxxxxxxxxxxxx. If that's what you need?

answered Jun 8, 2016 at 15:28

Praveen Kumar Purushothaman

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Comments

Jamiec · Accepted Answer · 2016-06-08 15:31:09Z

1

You really dont need regex for this, all you're doing is 3 substrings

string start to the mask start (1)
mask start to mask end (2)
mask end to string end. (3)

You then form the string back together by concatenating (1) above, the mask char to the length of (2) and finally (3).

var input = "1234567890";

var output = input.substring(0,4) + (Array(3).join('*')) + input.substring(6,10)

console.log(output)

answered Jun 8, 2016 at 15:31

Jamiec

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Comments

Kld · Accepted Answer · 2016-06-08 16:03:05Z

1

You can use capturing parentheses:

 "111122222333".replace( /(.{4})(.{5})(.*)/, '$1xxxxx$3');

1111 will be $1, replaced by itself.
22222 will be $2, replaced by xxxxx.
333 will be $3, replaced by itself.

edited Jun 8, 2016 at 16:03

answered Jun 8, 2016 at 15:56

Kld

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