CODEWITHSUNDEEP
javascriptregex
Amadeusz Blanik
Amadeusz Blanik

Reputation: 768

Regexp to match all the characters except spaces and 4 last characters

I need regexp for masking values except for the last 4 characters and spaces.

1234 4567 8901 2345 => **** **** **** 2345

123445678901 23 4 5 => ************ 23 4 5

[^\s](?=.{4,}$) doesn't pass for me

Upvotes: 3

Views: 460

Answers (3)

The fourth bird
The fourth bird

Reputation: 163277

You could update the pattern to assert 4 digits at the right until the end of the string.

\d(?=(?:[ \d]*\d){4}$)

Explanation

  • \d match a digit
  • (?= Positive lookahead
    • (?: Non capture group
      • [ \d]*\d Match 0+ times a space or digits followed by a single digit
    • ){4}$ and repeat that 4 times ans assert the end of the string
  • ) Close lookahead

Regex demo

[
  "1234 4567 8901 2345",
  "123445678901 23 4 5"
].forEach(s => console.log(s.replace(/\d(?=(?:[ \d]*\d){4}$)/gm, "*")));

Upvotes: 0

Christian Vincenzo Traina
Christian Vincenzo Traina

Reputation: 10384

I'm not sure you need a negative lookahead. If you have control over the input format then try this:

const number = '1234 5678 9012 3456';
const masked = number.replace(/\d{4}\s/g, '**** ');
console.log(masked);

If you want to accept also random spaces, try this:

const number = '1234 5678 9 012 3 45 6';
const masked = number.replace(/(\d\s*){4}\s/g, '**** ');
console.log(masked);

Upvotes: 0

Wiktor Stribiżew
Wiktor Stribiżew

Reputation: 626758

You can use

.replace(/\d(?=.*\d(?:\s*\d){3}\s*$)/g, '*')

See the regex demo

Regex details

  • \d - a digit
  • (?=.*\d(?:\s*\d){3}\s*$) - that is followed with any amount of any chars other than line break chars and then a digit, followed with three occurrences of 0+ whitespaces + a digit and then 0+ whitespaces at the end of string.

Upvotes: 1

Related Questions