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

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 "")

其他回答

请尝试以下代码

"250000".replace(/(\d)(?=(\d{3})+(?!\d))/g, '$1,');

答案:250000

这里是我见过的最好的JavaScript货币格式化程序:

Number.prototype.formatMoney = function(decPlaces, thouSeparator, decSeparator) {
    var n = this,
        decPlaces = isNaN(decPlaces = Math.abs(decPlaces)) ? 2 : decPlaces,
        decSeparator = decSeparator == undefined ? "." : decSeparator,
        thouSeparator = thouSeparator == undefined ? "," : thouSeparator,
        sign = n < 0 ? "-" : "",
        i = parseInt(n = Math.abs(+n || 0).toFixed(decPlaces)) + "",
        j = (j = i.length) > 3 ? j % 3 : 0;
    return sign + (j ? i.substr(0, j) + thouSeparator : "") + i.substr(j).replace(/(\d{3})(?=\d)/g, "$1" + thouSeparator) + (decPlaces ? decSeparator + Math.abs(n - i).toFixed(decPlaces).slice(2) : "");
};

它被重新格式化并从这里借用:如何将数字格式化为货币字符串

您必须提供您自己的货币代号(您使用了$以上)。

这样调用它(尽管注意,参数默认为2、逗号和句点,因此如果这是您的首选,则不需要提供任何参数):

var myMoney = 3543.75873;
var formattedMoney = '$' + myMoney.formatMoney(2, ',', '.'); // "$3,543.76"

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);
    }
}

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 "")

快速快捷的解决方案(适用于任何地方!)

(12345.67).toFixed(2).replace(/\d(?=(\d{3})+\.)/g, '$&,');  // 12,345.67

此解决方案背后的思想是用第一个匹配和逗号替换匹配的部分,即“$&,”。匹配是使用前瞻方法完成的。您可以将表达式读为“匹配一个数字,如果它后面跟着三个数字集(一个或多个)和一个点的序列”。

测验:

1        --> "1.00"
12       --> "12.00"
123      --> "123.00"
1234     --> "1,234.00"
12345    --> "12,345.00"
123456   --> "123,456.00"
1234567  --> "1,234,567.00"
12345.67 --> "12,345.67"

演示:http://jsfiddle.net/hAfMM/9571/


扩展短解决方案

您还可以扩展Number对象的原型,以添加对任意数量的小数[0..n]和数字组大小[0..x]的额外支持:

/**
 * Number.prototype.format(n, x)
 * 
 * @param integer n: length of decimal
 * @param integer x: length of sections
 */
Number.prototype.format = function(n, x) {
    var re = '\\d(?=(\\d{' + (x || 3) + '})+' + (n > 0 ? '\\.' : '$') + ')';
    return this.toFixed(Math.max(0, ~~n)).replace(new RegExp(re, 'g'), '$&,');
};

1234..format();           // "1,234"
12345..format(2);         // "12,345.00"
123456.7.format(3, 2);    // "12,34,56.700"
123456.789.format(2, 4);  // "12,3456.79"

演示/测试:http://jsfiddle.net/hAfMM/435/


超扩展短解决方案

在此超级扩展版本中,您可以设置不同的分隔符类型:

/**
 * Number.prototype.format(n, x, s, c)
 * 
 * @param integer n: length of decimal
 * @param integer x: length of whole part
 * @param mixed   s: sections delimiter
 * @param mixed   c: decimal delimiter
 */
Number.prototype.format = function(n, x, s, c) {
    var re = '\\d(?=(\\d{' + (x || 3) + '})+' + (n > 0 ? '\\D' : '$') + ')',
        num = this.toFixed(Math.max(0, ~~n));

    return (c ? num.replace('.', c) : num).replace(new RegExp(re, 'g'), '$&' + (s || ','));
};

12345678.9.format(2, 3, '.', ',');  // "12.345.678,90"
123456.789.format(4, 4, ' ', ':');  // "12 3456:7890"
12345678.9.format(0, 3, '-');       // "12-345-679"

演示/测试:http://jsfiddle.net/hAfMM/612/