function rnd(v,n=2) {
    return Math.round((v+Number.EPSILON)*Math.pow(10,n))/Math.pow(10,n)
}

这个很好地抓住了极端情况

es6接受答案版本:

function round(value, precision) {
    const multiplier = 10 ** (precision || 0);
    return Math.round(value * multiplier) / multiplier;
}

我找到了一个避免精度问题的方法:

function badRound (num, precision) {
    const x = 10 ** precision;
    return Math.round(num * x) / x
}
// badRound(1.005, 2) --> 1

function round (num, precision) {
    const x = 10 ** (precision + 1);
    const y = 10 ** precision;
    return Math.round(Math.round(num * x) / 10) / y
}
// round(1.005, 2) --> 1.01

如果你的源代码是typescript，你可以使用这样的函数:

public static ToFixedRounded(decimalNumber: number, fractionDigits: number): number {
    var rounded = Math.pow(10, fractionDigits);
    return (Math.round(decimalNumber * rounded) / rounded).toFixed(fractionDigits) as unknown as number;
}

我做了一个返回数字类型，也只在需要的时候放置小数(没有0填充)。

例子:

roundWithMaxPrecision(11.234, 2); //11.23
roundWithMaxPrecision(11.234, 1); //11.2
roundWithMaxPrecision(11.234, 4); //11.23
roundWithMaxPrecision(11.234, -1); //10

roundWithMaxPrecision(4.2, 2); //4.2
roundWithMaxPrecision(4.88, 1); //4.9

代码:

function roundWithMaxPrecision (n, precision) {
    const precisionWithPow10 = Math.pow(10, precision);
    return Math.round(n * precisionWithPow10) / precisionWithPow10;
}

使用toprecprecision方法:

var a = 1.2345
a.toPrecision(2)

// result "1.2"