一个更通用的N位舍入解决方案

function roundN(num,n){
  return parseFloat(Math.round(num * Math.pow(10, n)) /Math.pow(10,n)).toFixed(n);
}


console.log(roundN(1,2))
console.log(roundN(1.34,2))
console.log(roundN(1.35,2))
console.log(roundN(1.344,2))
console.log(roundN(1.345,2))
console.log(roundN(1.344,3))
console.log(roundN(1.345,3))
console.log(roundN(1.3444,3))
console.log(roundN(1.3455,3))

Output

1.00
1.34
1.35
1.34
1.35
1.344
1.345
1.344
1.346

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

试试下面的代码:

function numberWithCommas(number) { 

   var newval = parseFloat(Math.round(number * 100) / 100).toFixed(2);

   return newval.toString().replace(/\B(?=(\d{3})+(?!\d))/g, ",");
}

你可以试试下面的代码:

    function FormatNumber(number, numberOfDigits = 2) {
        try {
            return new Intl.NumberFormat('en-US').format(parseFloat(number).toFixed(numberOfDigits));
        } catch (error) {
            return 0;
        }
    }

    var test1 = FormatNumber('1000000.4444');
    alert(test1); // 1,000,000.44

    var test2 = FormatNumber(100000000000.55555555, 4);
    alert(test2); // 100,000,000,000.5556

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

这里有另一个只使用底价的四舍五入的解决方案，这意味着，确保计算的金额不会大于原始金额(有时交易需要):

Math.floor(num* 100 )/100;