Convert digits into words with JavaScript

Question

I am making a code which converts the given amount into words, heres is what I have got after googling. But I think its a little lengthy code to achieve a simple task. Two Regular Expressions and two for loops, I want something simpler.

I am trying to make it as shorter as possible. and will post what I come up with

Any suggestions?

var th = ['','thousand','million', 'billion','trillion'];
var dg = ['zero','one','two','three','four', 'five','six','seven','eight','nine'];
 var tn = ['ten','eleven','twelve','thirteen', 'fourteen','fifteen','sixteen', 'seventeen','eighteen','nineteen'];
 var tw = ['twenty','thirty','forty','fifty', 'sixty','seventy','eighty','ninety'];

function toWords(s) {
    s = s.toString();
    s = s.replace(/[\, ]/g,'');
    if (s != parseFloat(s)) return 'not a number';
    var x = s.indexOf('.');
    if (x == -1)
        x = s.length;
    if (x > 15)
        return 'too big';
    var n = s.split(''); 
    var str = '';
    var sk = 0;
    for (var i=0;   i < x;  i++) {
        if ((x-i)%3==2) { 
            if (n[i] == '1') {
                str += tn[Number(n[i+1])] + ' ';
                i++;
                sk=1;
            } else if (n[i]!=0) {
                str += tw[n[i]-2] + ' ';
                sk=1;
            }
        } else if (n[i]!=0) { // 0235
            str += dg[n[i]] +' ';
            if ((x-i)%3==0) str += 'hundred ';
            sk=1;
        }
        if ((x-i)%3==1) {
            if (sk)
                str += th[(x-i-1)/3] + ' ';
            sk=0;
        }
    }

    if (x != s.length) {
        var y = s.length;
        str += 'point ';
        for (var i=x+1; i<y; i++)
            str += dg[n[i]] +' ';
    }
    return str.replace(/\s+/g,' ');
}

Also, the above code converts to English numbering system like Million/Billion, I wan't South Asian numbering system. like in Lakhs and Crores

Answer 1

I like the result I got here which i think is easy to read and short enough to fit as a solution.

function NumInWords (number) {
  const first = ['','one ','two ','three ','four ', 'five ','six ','seven ','eight ','nine ','ten ','eleven ','twelve ','thirteen ','fourteen ','fifteen ','sixteen ','seventeen ','eighteen ','nineteen '];
  const tens = ['', '', 'twenty','thirty','forty','fifty', 'sixty','seventy','eighty','ninety'];
  const mad = ['', 'thousand', 'million', 'billion', 'trillion'];
  let word = '';

  for (let i = 0; i < mad.length; i++) {
    let tempNumber = number%(100*Math.pow(1000,i));
    if (Math.floor(tempNumber/Math.pow(1000,i)) !== 0) {
      if (Math.floor(tempNumber/Math.pow(1000,i)) < 20) {
        word = first[Math.floor(tempNumber/Math.pow(1000,i))] + mad[i] + ' ' + word;
      } else {
        word = tens[Math.floor(tempNumber/(10*Math.pow(1000,i)))] + '-' + first[Math.floor(tempNumber/Math.pow(1000,i))%10] + mad[i] + ' ' + word;
      }
    }

    tempNumber = number%(Math.pow(1000,i+1));
    if (Math.floor(tempNumber/(100*Math.pow(1000,i))) !== 0) word = first[Math.floor(tempNumber/(100*Math.pow(1000,i)))] + 'hunderd ' + word;
  }
    return word;
}

console.log(NumInWords(89754697976431))

And the result is :

eighty-nine trillion seven hundred fifty-four billion six hundred ninety-seven million nine hundred seventy-six thousand four hundred thirty-one

Answer 2

Try this code with a Turkish currency compliant JavaScript

function dene() {
         var inpt = document.getElementById("tar1").value;
         var spt = inpt.split('');
         spt.reverse();

         var tek = ["", "Bir", "İki", "Üç", "Dört", "Beş", "Altı", "Yedi", "Sekiz", "Dokuz"];
         var onlu = ["", "On", "Yirmi", "Otuz", "Kırk", "Elli", "Atmış", "Yetmiş", "Seksen", "Doksan"];
         var Yuz = ["", "Yüz", "İkiYüz", "Üçyüz", "DörtYüz", "BeşYüz", "AltıYüz", "YediYüz", "SekizYüz", "DokuzYüz"];
         var ska = ["", "", "", "", "Bin", "Milyon", "Milyar", "Trilyon", "Katrilyon", "Kentilyon"];
         var i, j;
         var bas3 = "";
         var bas6 = "";
         var bas9 = "";
         var bas12 = "";
         var total;

               for(i = 0; i < 1; i++) {

                      bas3 += Yuz[spt[i+2]] + onlu[spt[i+1]] + tek[spt[i]];
                      bas6 += Yuz[spt[i+5]] + onlu[spt[i+4]] + tek[spt[i+3]] + ska[4];
                      bas9 += Yuz[spt[i+8]] + onlu[spt[i+7]] + tek[spt[i+6]] + ska[5];
                      bas12 += Yuz[spt[i+11]] + onlu[spt[i+10]] + tek[spt[i+9]] + ska[6];


                   if(inpt.length < 4) {
                       bas6 = '';
                       bas9 = '';
                   }
                   if(inpt.length > 6 && inpt.slice(5, 6) == 0) {
                     bas6 = bas6.replace(/Bin/g, '');
                   }
                   if(inpt.length < 7) {
                       bas9 = '';
                   } 
                   if(inpt.length > 9 && inpt.slice(1,3) == 000){
                       bas9 = bas9.replace(/Milyon/g, '');
                   }

                   if(inpt.length < 10) {
                    bas12 = '';
                 }
             }

         total = bas12 + bas9 + bas6 + bas3;
         total = total.replace(NaN, '');
         total = total.replace(undefined, '');

        document.getElementById('demo').innerHTML = 
            total;


     }