Replace all chars with #, except for last 4

Question

function maskify(cc) {
    var dd = cc.toString();
    var hash = dd.replace((/./g), '#');
    for (var i = (hash.length - 4); i < hash.length; i++) {
        hash[i] = dd[i];
    }
    return hash;
}

I am trying to replace all chars with # except for last 4. Why isn't it working?

score 41 · Answer 1 · answered Feb 07 '16 at 17:34

41

You could do it like this:

dd.replace(/.(?=.{4,}$)/g, '#');

var dd = 'Hello dude';
var replaced = dd.replace(/.(?=.{4,}$)/g, '#');
document.write(replaced);

answered Feb 07 '16 at 17:34

MinusFour

13,090
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2

@LukasDavid, it basically replaces all characters where there's more than 4 characters afterwards. So when it get to 4 characters before the end of the string, the regex will fail and won't replace those characters. – MinusFour Feb 07 '16 at 18:22

score 4 · Answer 2 · answered Apr 04 '20 at 20:59

4

If you find the solution, try this trick

function maskify(cc) {
  return cc.slice(0, -4).replace(/./g, '#') + cc.slice(-4);
}

answered Apr 04 '20 at 20:59

krolovolk

426
4
16

score 0 · Accepted Answer · edited May 23 '17 at 12:33

0

To replace a character in a string at a given index, hash[i] = dd[i] doesn't work. Strings are immutable in Javascript. See How do I replace a character at a particular index in JavaScript? for some advice on that.

edited May 23 '17 at 12:33

Community

1
1

answered Feb 07 '16 at 17:33

Adi Levin

5,025
1
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25

Replace all chars with #, except for last 4

3 Answers3