我想用JavaScript格式化价格。我想要一个函数,它将浮点作为参数,并返回如下格式的字符串:

"$ 2,500.00"

我该怎么做?


当前回答

String.prototype.toPrice = function () {
    var v;
    if (/^\d+(,\d+)$/.test(this))
        v = this.replace(/,/, '.');
    else if (/^\d+((,\d{3})*(\.\d+)?)?$/.test(this))
        v = this.replace(/,/g, "");
    else if (/^\d+((.\d{3})*(,\d+)?)?$/.test(this))
        v = this.replace(/\./g, "").replace(/,/, ".");
    var x = parseFloat(v).toFixed(2).toString().split("."),
    x1 = x[0],
    x2 = ((x.length == 2) ? "." + x[1] : ".00"),
    exp = /^([0-9]+)(\d{3})/;
    while (exp.test(x1))
        x1 = x1.replace(exp, "$1" + "," + "$2");
    return x1 + x2;
}

alert("123123".toPrice()); //123,123.00
alert("123123,316".toPrice()); //123,123.32
alert("12,312,313.33213".toPrice()); //12,312,313.33
alert("123.312.321,32132".toPrice()); //123,312,321.32

其他回答

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"

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);
    }
}

这可能奏效:

function format_currency(v, number_of_decimals, decimal_separator, currency_sign){
  return (isNaN(v)? v : currency_sign + parseInt(v||0).toLocaleString() + decimal_separator + (v*1).toFixed(number_of_decimals).slice(-number_of_decimals));
}

没有循环,没有正则表达式,没有数组,没有奇异条件。

一种仅满足原始要求的极简方法:

function formatMoney(n) {
    return "$ " + (Math.round(n * 100) / 100).toLocaleString();
}

@丹尼尔·马格廖拉:你说得对。以上是一个仓促而不完整的实施。以下是正确的实施方式:

function formatMoney(n) {
    return "$ " + n.toLocaleString().split(".")[0] + "."
        + n.toFixed(2).split(".")[1];
}

这是我的。。。

function thousandCommas(num) {
  num = num.toString().split('.');
  var ints = num[0].split('').reverse();
  for (var out=[],len=ints.length,i=0; i < len; i++) {
    if (i > 0 && (i % 3) === 0) out.push(',');
    out.push(ints[i]);
  }
  out = out.reverse() && out.join('');
  if (num.length === 2) out += '.' + num[1];
  return out;
}