function currencyFormat (num) {
    return "$" + num.toFixed(2).replace(/(\d)(?=(\d{3})+(?!\d))/g, "$1,")
}

console.info(currencyFormat(2665));   // $2,665.00
console.info(currencyFormat(102665)); // $102,665.00

刚刚遇到了这个最长的线程，下面是我的解决方案:

parseFloat(Math.round((parseFloat(num * 100)).toFixed(2)) / 100 ).toFixed(2)

如果有人能找到漏洞就告诉我

将数字转换为字符串，只保留两个小数:

var num = 5.56789;
var n = num.toFixed(2);

n的结果是:

5.57

我建议你用

new Intl.NumberFormat('de-DE', { minimumFractionDigits: 2, maximumFractionDigits: 2 }).format(num)

这样，您还将拥有指定国家的本地格式，并且它将确保显示精确的2个小数(无论num是1还是1.12345，它将分别显示1.00和1.12)

在这个例子中，我使用德语本地化，因为我想我的数字显示与千分隔符，所以这将是一些输出:

1 => 1,00
1.12 => 1,12
1.1234 => 1,12
1234 => 1.234,00
1234.1234 => 1.234,12

RegExp -替代方法

在输入时，你有字符串(因为你使用解析)，所以我们可以通过只使用字符串操作和整数计算得到结果

let toFix2 = (n) = > n.replace (d / \(?) -(+)。\ \ \ \ d (d + 1) / (_), s, i、d (r) = {> 让k= (+r[0]>=5)+ +d - (r==5 && s=='-'); 返回s +(+i+(k>99)) + "。"(k +(> 99%) ? ? 9: 00”(k≥0 " k " + k)); }) / /测试 console.log toFix2(“1”)); console.log (toFix2 1.341”()); console.log (toFix2 1.345”()); console.log (toFix2 1.005”());

解释

s is sign, i is integer part, d are first two digits after dot, r are other digits (we use r[0] value to calc rounding) k contains information about last two digits (represented as integer number) if r[0] is >=5 then we add 1 to d - but in case when we have minus number (s=='-') and r is exact equal to 5 then in this case we substract 1 (for compatibility reasons - in same way Math.round works for minus numbers e.g Math.round(-1.5)==-1) after that if last two digits k are greater than 99 then we add one to integer part i

好消息! ! 似乎javascript的新版本ES2020(我只是使用)提供了这个函数的新行为。

let ff:number =3
console.info(ff.toFixed(2)) //3.00

根据需要。