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。

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;
}

主要部分是插入千个分隔符，可以这样做：

<script type="text/javascript">
  function ins1000Sep(val) {
    val = val.split(".");
    val[0] = val[0].split("").reverse().join("");
    val[0] = val[0].replace(/(\d{3})/g, "$1,");
    val[0] = val[0].split("").reverse().join("");
    val[0] = val[0].indexOf(",") == 0 ? val[0].substring(1) : val[0];
    return val.join(".");
  }

  function rem1000Sep(val) {
    return val.replace(/,/g, "");
  }

  function formatNum(val) {
    val = Math.round(val*100)/100;
    val = ("" + val).indexOf(".") > -1 ? val + "00" : val + ".00";
    var dec = val.indexOf(".");
    return dec == val.length-3 || dec == 0 ? val : val.substring(0, dec+3);
  }
</script>

<button onclick="alert(ins1000Sep(formatNum(12313231)));">

乔纳森·M的代码对我来说太复杂了，所以我重写了它，在Firefox v30上获得了30%的速度提升，在Chrome v35上获得了60%的速度提升(http://jsperf.com/number-formating2):

Number.prototype.formatNumber = function(decPlaces, thouSeparator, decSeparator) {
    decPlaces = isNaN(decPlaces = Math.abs(decPlaces)) ? 2 : decPlaces;
    decSeparator = decSeparator == undefined ? "." : decSeparator;
    thouSeparator = thouSeparator == undefined ? "," : thouSeparator;

    var n = this.toFixed(decPlaces);
    if (decPlaces) {
        var i = n.substr(0, n.length - (decPlaces + 1));
        var j = decSeparator + n.substr(-decPlaces);
    } else {
        i = n;
        j = '';
    }

    function reverse(str) {
        var sr = '';
        for (var l = str.length - 1; l >= 0; l--) {
            sr += str.charAt(l);
        }
        return sr;
    }

    if (parseInt(i)) {
        i = reverse(reverse(i).replace(/(\d{3})(?=\d)/g, "$1" + thouSeparator));
    }
    return i + j;
};

用法：

var sum = 123456789.5698;
var formatted = '$' + sum.formatNumber(2, ',', '.'); // "$123,456,789.57"

适用于所有当前浏览器

使用toLocaleString以货币的语言敏感表示形式格式化货币（使用ISO 4217货币代码）。

(2500).toLocaleString("en-GB", {style: "currency", currency: "GBP", minimumFractionDigits: 2})

avenmore的南非兰特代码片段示例：

console.log（（2500）.toLocaleString（“en-ZA”，｛style:“currency”，currency:“ZAR”，minimumFractionDigits:2｝））//->2 500,00兰特console.log（（2500）.toLocaleString（“en-GB”，｛style:“currency”，currency:“ZAR”，minimumFractionDigits:2｝））//->2500欧元

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 "")