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 R
For those who are looking for imperial/english naming conventions.
Based on @Salman's answer
var a = ['','one ','two ','three ','four ', 'five ','six ','seven ','eight ','nine ','ten ','eleven ','twelve ','thirteen ','fourteen ','fifteen ','sixteen ','seventeen ','eighteen ','nineteen '];
var b = ['', '', 'twenty','thirty','forty','fifty', 'sixty','seventy','eighty','ninety'];
function inWords (num) {
if ((num = num.toString()).length > 12) return 'overflow';
n = ('00000000000' + num).substr(-12).match(/^(\d{3})(\d{3})(\d{3})(\d{1})(\d{2})$/);
if (!n) return; var str = '';
str += (n[1] != 0) ? (Number(n[1]) > 99 ? this.a[Number(n[1][0])] + 'hundred ' : '') + (a[Number(n[1])] || b[n[1][1]] + ' ' + a[n[1][2]]) + 'billion ' : '';
str += (n[2] != 0) ? (Number(n[2]) > 99 ? this.a[Number(n[2][0])] + 'hundred ' : '') + (a[Number(n[2])] || b[n[2][1]] + ' ' + a[n[2][2]]) + 'million ' : '';
str += (n[3] != 0) ? (Number(n[3]) > 99 ? this.a[Number(n[3][0])] + 'hundred ' : '') + (a[Number(n[3])] || b[n[3][1]] + ' ' + a[n[3][2]]) + 'thousand ' : '';
str += (n[4] != 0) ? (a[Number(n[4])] || b[n[4][0]] + ' ' + a[n[4][1]]) + 'hundred ' : '';
str += (Number(n[5]) !== 0) ? ((str !== '') ? 'and ' : '') +
(this.a[Number(n[5])] || this.b[n[5][0]] + ' ' +
this.a[n[5][1]]) + '' : '';
return str;
}
document.getElementById('number').onkeyup = function () {
document.getElementById('words').innerHTML = inWords(document.getElementById('number').value);
};