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

这是用的安全解。这将修复。15的舍入

function myFunction(a) {

    return Math.round((a + Number.EPSILON) * 100) / 100
}

你是这个意思吗?

@razu提供的答案是最好的。这是一个稍微重构的版本。

代码片段仍然不能返回类似showAsFloat(2.3346)的正确值(结果2.33，但应该是2.34)。所以，也看。

const showAsFloat = (input, decimals = 2, asString = false) => { if (input === null || input.constructor === Boolean || isNaN(+input)) { return input; } const converted = +( `${Math.round( parseFloat( `${input}e${decimals}` ) )}e-${decimals}` ); return asString ? converted.toFixed(decimals) : converted }; document.querySelector('#result').textContent = [ 'command | result', '-----------------------------------------------', 'showAsFloat(1); | ' + showAsFloat(1), 'showAsFloat(1.314); | ' + showAsFloat(1.314), 'showAsFloat(\'notanumber\') | ' + showAsFloat('notanumber'), 'showAsFloat(\'23.44567\', 3) | ' + showAsFloat('23.44567', 3), 'showAsFloat(2456198, 5, true)| ' + showAsFloat('24568', 5, true), 'showAsFloat(2456198, 5) | ' + showAsFloat('24568', 5), 'showAsFloat(0, 2, true); | ' + showAsFloat(0, 2, true), 'showAsFloat(1.345); | ' + showAsFloat(1.345), 'showAsFloat(0.005); | ' + showAsFloat(0.005), 'showAsFloat(null); | ' + showAsFloat(null), ].join('\n'); <pre id="result"></pre>

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。

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

}