Phone number format javascript

Question

I'm trying to format some phone numbers with regexp. The phone numbers i have stored are of diffrent length and all phone numbers have their country codes. Lets say a number looks like this: 00460708186681. I want that one to be formatted as: +46 (0)708-18 66 81. The first three groups are easy:

/00([\d]{2})([\d]{1})([\d]{3})/

Because the beginning of the number will always be the same. It's the last part where i dont know the length of the remaining chars (and i want them to be divided into groups of two).

Why are you doing this with a regex? It's not the right tool for this job, even if it can do it. It will be much faster, and probably simpler to just use fixed substrings. — Ariel
– Ariel, Commented Jul 17, 2012 at 9:39

Vala · Accepted Answer · 2012-07-17 09:39:37Z

1

Start with the following regex as your base:

00(\d{2})(\d)(\d{3})(\d{2})

Now add another pair (\d{2})? (note the '?') for every pair it's possible to have. So if the maximum number of pairs is 3, you do the following:

00(\d{2})(\d)(\d{3})(\d{2})(\d{2})?(\d{2})?

It's not pretty, but you need to do it that way in order to get them grouped correctly. Or you could simply do:

00(\d{2})(\d)(\d{3})((?:\d{2}){2,4}) // In this case it's 2 to 4 pairs of digits on the end.

To match the string, then you can manually add spaces in the last group which will contain all the digits after your initial formatting.

answered Jul 17, 2012 at 9:39

Vala

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Bergi · Accepted Answer · 2012-07-17 09:53:45Z

1

Regex is not the best solution here. Use this:

var num = new String("00460708186681"),
    i = 2,
    p,
    res = "+"+num.substr(i,2)+" ("+num.charAt(i+=2)+")"+num.substr(++i,3)+"-"+num.substr(i+=3,2);
while (p=num.substr(i+=2,2))
    res+= "\u00A0"+p;
return res;

Which will work for infinitely long numbers. A regex can't do that.

answered Jul 17, 2012 at 9:53

Bergi

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1 Comment

Vala Over a year ago

He's dealing with phone numbers. They might be of variable length, but there's certainly a well known maximum (and it's not large).

gaurang171 · Accepted Answer · 2012-07-17 13:23:57Z

0

Here is solution :

Javascript:

var num="00460708186681";
var res=num.replace(/([0+]{2})([1-9]{2})([\d]{1})([\d]{3})([\d]{2})([\d]{2})([\d]{2})/g,'+$2($3)-$4 $5 $6 $7');
alert(res);

Try above script on http://codebins.com/

answered Jul 17, 2012 at 13:23

gaurang171

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1 Comment

Vala Over a year ago

[0+]{2} allows any number of 0s (1 or more 0s, twice). The second part [1-9]{2} will not match 0010 for example which should be considered valid, you could use [1-9][0-9] instead. Also, there's no need to put single values inside [] (you've done it with 0+ and all the \ds) as that's an OR in regex.

Leonel Kahameni · Accepted Answer · 2020-09-26 12:08:00Z

Here is an example of code done in TypeScript, but you can surely parse it into JavaScript:

const formatPhoneNumber = (phoneNumber: number, format?: string, code?: boolean): string => {
  let _format = format === undefined ? '---  ---  ---' : format
  let x = 1
  const newFormat = _format.replace(/-+/g, () => '$' + x++)
  let regex = '^'
  _format.match(/-+/g)?.forEach((c, k) => {
      regex += `(\\d{${c.length}})?`
  })
  return (code !== undefined && code === true ? '+' : '') + (phoneNumber.toString().concat(_format.split(' ').join(''))).slice(0, _format.split(' ').join('').length).toString().replace(new RegExp(regex, 'g'), newFormat)}

To test it, you can do:

const formatedPhoneNumber = formatPhoneNumber(600000000) // => 600  000 000
const formatedPhoneNumberWithCountryCode = formatPhoneNumber(237600000000, 'xxx xxx xxx xxx', true) // => +237 600 000 000

This example is perfectly modular, so you can do specify your own format (template) to have a different output!

Note: I'm Cameroonian, that's why I used that strange template, Feel free to adapt the template to the needed phone number template

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