数字（值）.to固定（2）.replace（/（\d）（？=（\d｛3｝）+（？！\d））/g，“$1，”）

我主要基于VisionN的回答：

function format (val) {
  val = (+val).toLocaleString();
  val = (+val).toFixed(2);
  val += "";
  return val.replace(/(\d)(?=(\d{3})+(?:\.\d+)?$)/g, "$1" + format.thousands);
}

(function (isUS) {
  format.decimal =   isUS ? "." : ",";
  format.thousands = isUS ? "," : ".";
}(("" + (+(0.00).toLocaleString()).toFixed(2)).indexOf(".") > 0));

我用输入进行了测试：

[   ""
  , "1"
  , "12"
  , "123"
  , "1234"
  , "12345"
  , "123456"
  , "1234567"
  , "12345678"
  , "123456789"
  , "1234567890"
  , ".12"
  , "1.12"
  , "12.12"
  , "123.12"
  , "1234.12"
  , "12345.12"
  , "123456.12"
  , "1234567.12"
  , "12345678.12"
  , "123456789.12"
  , "1234567890.12"
  , "1234567890.123"
  , "1234567890.125"
].forEach(function (item) {
  console.log(format(item));
});

得到了这些结果：

0.00
1.00
12.00
123.00
1,234.00
12,345.00
123,456.00
1,234,567.00
12,345,678.00
123,456,789.00
1,234,567,890.00
0.12
1.12
12.12
123.12
1,234.12
12,345.12
123,456.12
1,234,567.12
12,345,678.12
123,456,789.12
1,234,567,890.12
1,234,567,890.12
1,234,567,890.13

只是为了好玩。

Patrick Desjardins（前Daok）的例子对我很有用。如果有人感兴趣，我将其移植到CoffeeScript。

Number.prototype.toMoney = (decimals = 2, decimal_separator = ".", thousands_separator = ",") ->
    n = this
    c = if isNaN(decimals) then 2 else Math.abs decimals
    sign = if n < 0 then "-" else ""
    i = parseInt(n = Math.abs(n).toFixed(c)) + ''
    j = if (j = i.length) > 3 then j % 3 else 0
    x = if j then i.substr(0, j) + thousands_separator else ''
    y = i.substr(j).replace(/(\d{3})(?=\d)/g, "$1" + thousands_separator)
    z = if c then decimal_separator + Math.abs(n - i).toFixed(c).slice(2) else ''
    sign + x + y + z

Patrick热门答案的CoffeeScript：

Number::formatMoney = (decimalPlaces, decimalChar, thousandsChar) ->
  n = this
  c = decimalPlaces
  d = decimalChar
  t = thousandsChar
  c = (if isNaN(c = Math.abs(c)) then 2 else c)
  d = (if d is undefined then "." else d)
  t = (if t is undefined then "," else t)
  s = (if n < 0 then "-" else "")
  i = parseInt(n = Math.abs(+n or 0).toFixed(c)) + ""
  j = (if (j = i.length) > 3 then j % 3 else 0)
  s + (if j then i.substr(0, j) + t else "") + i.substr(j).replace(/(\d{3})(?=\d)/g, "$1" + t) + (if c then d + Math.abs(n - i).toFixed(c).slice(2) else "")

使用regexp的更快方法：

Number.prototype.toMonetaryString = function() {
  var n = this.toFixed(2), m;
  //var = this.toFixed(2).replace(/\./, ','); For comma separator
  // with a space for thousands separator
  while ((m = n.replace(/(\d)(\d\d\d)\b/g, '$1 $2')) != n)
      n = m;
  return m;
}

String.prototype.fromMonetaryToNumber = function(s) {
  return this.replace(/[^\d-]+/g, '')/100;
}

通常，有多种方法可以执行相同的操作，但我会避免使用Number.prototype.toLocaleString，因为它可以根据用户设置返回不同的值。

我也不建议扩展Number.prototype-扩展原生对象原型是一种糟糕的做法，因为它可能会与其他人的代码（例如库/框架/插件）发生冲突，并且可能与未来的JavaScript实现/版本不兼容。

我认为正则表达式是解决这个问题的最佳方法，下面是我的实现：

/**
 * Converts number into currency format
 * @param {number} number    Number that should be converted.
 * @param {string} [decimalSeparator]    Decimal separator, defaults to '.'.
 * @param {string} [thousandsSeparator]    Thousands separator, defaults to ','.
 * @param {int} [nDecimalDigits]    Number of decimal digits, defaults to `2`.
 * @return {string} Formatted string (e.g. numberToCurrency(12345.67) returns '12,345.67')
 */
function numberToCurrency(number, decimalSeparator, thousandsSeparator, nDecimalDigits){
    //default values
    decimalSeparator = decimalSeparator || '.';
    thousandsSeparator = thousandsSeparator || ',';
    nDecimalDigits = nDecimalDigits == null? 2 : nDecimalDigits;

    var fixed = number.toFixed(nDecimalDigits), //limit/add decimal digits
        parts = new RegExp('^(-?\\d{1,3})((?:\\d{3})+)(\\.(\\d{'+ nDecimalDigits +'}))?$').exec( fixed ); //separate begin [$1], middle [$2] and decimal digits [$4]

    if(parts){ //number >= 1000 || number <= -1000
        return parts[1] + parts[2].replace(/\d{3}/g, thousandsSeparator + '$&') + (parts[4] ? decimalSeparator + parts[4] : '');
    }else{
        return fixed.replace('.', decimalSeparator);
    }
}