：看起来这比我想象的要有用。我刚刚在npm上发布了此内容。https://www.npmjs.com/package/num- words
这是一个较短的代码。一个RegEx，没有循环。在南亚编号系统中根据需要进行转换
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 > 9) return 'overflow';
n = ('000000000' + num).substr(-9).match(/^(\d{2})(\d{2})(\d{2})(\d{1})(\d{2})$/);
if (!n) return; var str = '';
str += (n[1] != 0) ? (a[Number(n[1])] || b[n[1][0]] + ' ' + a[n[1][1]]) + 'crore ' : '';
str += (n[2] != 0) ? (a[Number(n[2])] || b[n[2][0]] + ' ' + a[n[2][1]]) + 'lakh ' : '';
str += (n[3] != 0) ? (a[Number(n[3])] || b[n[3][0]] + ' ' + a[n[3][1]]) + 'thousand ' : '';
str += (n[4] != 0) ? (a[Number(n[4])] || b[n[4][0]] + ' ' + a[n[4][1]]) + 'hundred ' : '';
str += (n[5] != 0) ? ((str != '') ? 'and ' : '') + (a[Number(n[5])] || b[n[5][0]] + ' ' + a[n[5][1]]) + 'only ' : '';
return str;
}
document.getElementById('number').onkeyup = function () {
document.getElementById('words').innerHTML = inWords(document.getElementById('number').value);
};
<span id="words"></span>
<input id="number" type="text" />
唯一的限制是，您最多可以转换9位数字，我认为在大多数情况下，这已经足够了。
602392714
清零编程群