var num = new Number(14.12); console.log (num.toPrecision (2));/ /输出14 console.log (num.toPrecision (3));/ /输出14.1 console.log (num.toPrecision (4));/ /输出14.12 console.log (num.toPrecision (5));/ /输出14.120

var num1 = “0.1”; document.getElementById（'num1'）.innerHTML =（Math.round（num1 * 100） / 100）.toFixed（2）; var num2 = “1.341”; document.getElementById（'num2'）.innerHTML =（Math.round（num2 * 100） / 100）.toFixed（2）; var num3 = “1.345”; document.getElementById（'num3'）.innerHTML =（Math.round（num3 * 100） / 100）.toFixed（2）; 跨度 { 边框：1px 实心 #000; 边距：5px; 填充：5px; } <span id=“num1”></span> <span id=“num2”></span> <span id=“num3”></span>

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

function formatValeurDecimal(valeurAFormate,longueurPartieEntier,longueurPartieDecimal){

valeurAFormate = valeurAFormate.replace(",",".")
valeurAFormate = parseFloat(valeurAFormate).toFixed(longueurPartieDecimal)
if(valeurAFormate == 'NaN'){
    return 0
}

//____________________valeurPartieEntier__________________________________
var valeurPartieEntier = valeurAFormate | 0

var strValeur = valeurPartieEntier.toString()
strValeur = strValeur.substring(0, longueurPartieEntier)
valeurPartieEntier = strValeur

//____________________valeurPartieDecimal__________________________________
strValeur = valeurAFormate
strValeur = strValeur.substring(strValeur.indexOf('.')+1)
var valeurPartieDecimal = strValeur

valeurAFormate = valeurPartieEntier +'.'+valeurPartieDecimal
if(valeurAFormate == null){
    valeurAFormate = 0
}

return valeurAFormate

}

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

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