Masking phone number with regex in javascript

Question

My application has a specific phone number format which looks like 999.111.222, which I have a regex pattern to mask it on front-end:

/[0-9]{3}\.[0-9]{3}\.([0-9]{3})/

But recently, the format was changed to allow the middle three digits to have one less digit, so now both 999.11.222 and 999.111.222 match. How can I change my regex accordingly?

"999.111.222".replace(/[0-9]{3}\.[0-9]{3}\.([0-9]{3})/, '<div>xxx.xxx.$1</div>')

expected output:

"999.111.222" // xxx.xxx.222
"999.11.222" // xxx.xx.222

MofX · Accepted Answer · 2019-08-07 13:01:45Z

5

Replace {3} with {2,3} to match two or three digits.

/[0-9]{3}\.[0-9]{2,3}\.([0-9]{3})/

For reference see e.g. MDN

edited Aug 7, 2019 at 13:01

MofX

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answered Aug 7, 2019 at 12:58

Aedvald Tseh

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1 Comment

Wiktor Stribiżew Over a year ago

That will not help OP, since the second part (between dots) needs to be masked with x.

Wiktor Stribiżew · Accepted Answer · 2019-08-07 14:05:21Z

2

Use

console.log(
  "999.11.222".replace(/[0-9]{3}\.([0-9]{2,3})\.([0-9]{3})/, function ($0, $1, $2)
  { return '<div>xxx.' + $1.replace(/\d/g, 'x') + '.' + $2 + '</div>'; })
)

The ([0-9]{2,3}) first capturing group will match 2 or 3 digits, and in the callback method used as the replacement argument, all the digits from th first group are replaced with x.

You may further customize the pattern for the first set of digits, too.

edited Aug 7, 2019 at 14:05

answered Aug 7, 2019 at 12:59

Wiktor Stribiżew

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Comments

guijob · Accepted Answer · 2019-08-07 13:08:56Z

1

In fact, you should change not only your regex but also your callback replace function:

const regex = /[0-9]{3}\.([0-9]{2,3})\.([0-9]{3})/;
const cbFn = (all, g1, g2) =>`<div>xxx.xx${(g1.length === 3 ? 'x' : '')}.${g2}</div>`;

const a = "999.11.222".replace(regex, cbFn);
const b = "999.111.222".replace(regex, cbFn);

console.log(a, b);

To change regex you could add a term with {2,3} quantifier, as already suggested, and create a new group. Then, in replace cb function, you can use length to know if you must put a new x.

edited Aug 7, 2019 at 13:08

answered Aug 7, 2019 at 12:57

guijob

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