Format string number as currency

Question

I would like a mask function to format number (as string) like a lot of mask plugins do.

Desired behavior:

format('1') // '0,01'
format('12') // '0,12'
format('123') // '1,23'
format('12345678') // '123.456,78'

It would be better if the formatter achieve this using only regex functions.

[EDIT]

A searched a lot of questions here in stackoverflow but mostly answers consist in add two decimal cases at the end of string. That's not what I expected.

EXPECTED:

format('1') // '0,01'
format('123') // '1,23'

NOT EXPECTED:

format('1') // '1,00'
format('123') // '123,00'

Answer 1

Here is a two-step solution: get the digits sorted into 3+ digit numbers and 1-2 digit numbers with

^(?:(\d+)(\d{2})|(\d{1,2}))$

See the regex demo

So, the numbers bigger than 1 will have their integer parts in Group 1 and decimal parts in Group 2, and Group 3 will hold all fractional numbers less than 0.

Then, we check the number we get and only add thousand separators to the integer parts with

.replace(/(\d)(?=(\d{3})+(?!\d))/g, "$1.")

See the JS demo:

function format(s) {
  var re = /^(?:(\d+)(\d{2})|(\d{1,2}))$/; 
  if ((m = s.match(re)) !== null) {   
      if (m[3]) {                     // We have 1 or 2 digit number
          return m[3].length == 1 ? 
              "0,0" + m[3] : "0," + m[3];
      } else {                       // We have a longer number
          return m[1].replace(/(\d)(?=(\d{3})+(?!\d))/g, "$1.") + "," + m[2];
      }
    
    }
}

document.body.innerHTML = format('1') + "<br/>"; // '0,01'
document.body.innerHTML += format('12') + "<br/>"; // '0,12'
document.body.innerHTML += format('123') + "<br/>"; // '1,23'
document.body.innerHTML += format('12345678') + "<br/>"; // '123.456,78'

Actually, you may add the thousand and decimal separators as the function arguments:

function format(str, thd, dec) {
  var re = /^(?:(\d+)(\d{2})|(\d{1,2}))$/; 
  if ((m = str.match(re)) !== null) {   
    if (m[3]) {                     // We have 1 or 2 digit number
      return m[3].length == 1 ? 
         "0" + dec + "0" + m[3] : "0" + dec + m[3];
    } else {                       // We have a longer number
         return m[1].replace(/(\d)(?=(\d{3})+(?!\d))/g, "$1" + thd) + dec + m[2];
    }
  }
}

var thd = "."; // Thousand separator
var dec = ","; // Decimal separator
document.body.innerHTML = format('1', thd, dec) + "<br/>"; // '0,01'
document.body.innerHTML += format('12', thd, dec) + "<br/>"; // '0,12'
document.body.innerHTML += format('123', thd, dec) + "<br/>"; // '1,23'
document.body.innerHTML += format('12345678', thd, dec) + "<br/>"; // '123.456,78'