避免数学不精确的最好方法是转换为字符串，但要确保它是你期望的“点”格式，使用toLocaleString:

函数getDecimals(n) { //注意，maximumSignificantDigits默认为3，所以如果不更改，您的小数将被四舍五入。 const parts = n.toLocaleString('en-US'， {maximumSignificantDigits: 18}).split('.') 返回的部分。长度> 1 ?数量(' 0。' +部分[1]):0 ｝ console.log (getDecimals (10.58))

你可以简单地使用parseInt()函数来帮助，例如: 令decimal = 3.2; let remainder = decimal - parseInt(decimal); document . write(剩余部分);

你可以将其转换为字符串，并使用replace方法将整数部分替换为零，然后将结果转换回一个数字:

var number = 123.123812,
    decimals = +number.toString().replace(/^[^\.]+/,'0');

为什么0.2999999999999998是一个可以接受的答案?如果我是提问者，我希望答案是。3。我们在这里得到的是错误的精度，我对floor、%等的实验表明，Javascript对这些操作喜欢错误的精度。所以我认为使用字符串转换的答案是正确的。

我会这样做:

var decPart = (n+"").split(".")[1];

具体来说，我使用100233.1，我想要答案“.1”。

2021年更新

优化版本处理精度(或不)。

// Global variables. const DEFAULT_PRECISION = 16; const MAX_CACHED_PRECISION = 20; // Helper function to avoid numerical imprecision from Math.pow(10, x). const _pow10 = p => parseFloat(`1e+${p}`); // Cache precision coefficients, up to a precision of 20 decimal digits. const PRECISION_COEFS = new Array(MAX_CACHED_PRECISION); for (let i = 0; i !== MAX_CACHED_PRECISION; ++i) { PRECISION_COEFS[i] = _pow10(i); } // Function to get a power of 10 coefficient, // optimized for both speed and precision. const pow10 = p => PRECISION_COEFS[p] || _pow10(p); // Function to trunc a positive number, optimized for speed. // See: https://stackoverflow.com/questions/38702724/math-floor-vs-math-trunc-javascript const trunc = v => (v < 1e8 && ~~v) || Math.trunc(v); // Helper function to get the decimal part when the number is positive, // optimized for speed. // Note: caching 1 / c or 1e-precision still leads to numerical errors. // So we have to pay the price of the division by c. const _getDecimals = (v = 0, precision = DEFAULT_PRECISION) => { const c = pow10(precision); // Get precision coef. const i = trunc(v); // Get integer. const d = v - i; // Get decimal. return Math.round(d * c) / c; } // Augmenting Number proto. Number.prototype.getDecimals = function(precision) { return (isFinite(this) && (precision ? ( (this < 0 && -_getDecimals(-this, precision)) || _getDecimals(this, precision) ) : this % 1)) || 0; } // Independent function. const getDecimals = (input, precision) => (isFinite(input) && ( precision ? ( (this < 0 && -_getDecimals(-this, precision)) || _getDecimals(this, precision) ) : this % 1 )) || 0; // Tests: const test = (value, precision) => ( console.log(value, '|', precision, '-->', value.getDecimals(precision)) ); test(1.001 % 1); // --> 0.0009999999999998899 test(1.001 % 1, 16); // --> 0.000999999999999 test(1.001 % 1, 15); // --> 0.001 test(1.001 % 1, 3); // --> 0.001 test(1.001 % 1, 2); // --> 0 test(-1.001 % 1, 16); // --> -0.000999999999999 test(-1.001 % 1, 15); // --> -0.001 test(-1.001 % 1, 3); // --> -0.001 test(-1.001 % 1, 2); // --> 0

浮点小数点符号和数字格式可以依赖于国家(.，)，因此保留浮点部分的独立解为:

getFloatDecimalPortion = function(x) {
    x = Math.abs(parseFloat(x));
    let n = parseInt(x);
    return Number((x - n).toFixed(Math.abs((""+x).length - (""+n).length - 1)));
}

-它是国际化的解决方案，而不是位置依赖:

getFloatDecimalPortion = x => parseFloat("0." + ((x + "").split(".")[1]));

方案描述一步步:

parseFloat() for guaranteeing input cocrrection Math.abs() for avoiding problems with negative numbers n = parseInt(x) for getting decimal part x - n for substracting decimal part We have now number with zero decimal part, but JavaScript could give us additional floating part digits, which we do not want So, limit additional digits by calling toFixed() with count of digits in floating part of original float number x. Count is calculated as difference between length of original number x and number n in their string representation.