我需要显示一个货币值的格式1K等于一千，或1.1K, 1.2K, 1.9K等，如果它不是一个偶数千，否则如果低于一千，显示正常500,100,250等，使用JavaScript格式化的数字?
当前回答
ES2020在Intl中增加了对此的支持。使用如下表示法:
let formatter = Intl。NumberFormat('en'，{符号:'紧凑'}); //示例1 让million = formatter.format(1e6); //示例2 Let billion = formatter.format(1e9); / /打印 console.log(million == '1M'， billion == '1B');
注意如上所示，第二个示例生成1B而不是1G。 NumberFormat规格:
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Intl/NumberFormat/NumberFormat https://tc39.es/ecma402#numberformat-objects
注意，目前并不是所有的浏览器都支持ES2020，所以你可能需要这个 Polyfill: https://formatjs.io/docs/polyfills/intl-numberformat
其他回答
这个函数可以将巨大的数字(正数和负数)转换为读者友好的格式，而不会失去其精度:
function abbrNum(n) { if (!n || (n && typeof n !== 'number')) { return ''; } const ranges = [ { divider: 1e12 , suffix: 't' }, { divider: 1e9 , suffix: 'b' }, { divider: 1e6 , suffix: 'm' }, { divider: 1e3 , suffix: 'k' } ]; const range = ranges.find(r => Math.abs(n) >= r.divider); if (range) { return (n / range.divider).toString() + range.suffix; } return n.toString(); } /* test cases */ let testAry = [99, 1200, -150000, 9000000]; let resultAry = testAry.map(abbrNum); console.log("result array: " + resultAry);
这是非常优雅的。
function formatToUnits(number, precision) {
const abbrev = ['', 'k', 'm', 'b', 't'];
const unrangifiedOrder = Math.floor(Math.log10(Math.abs(number)) / 3)
const order = Math.max(0, Math.min(unrangifiedOrder, abbrev.length -1 ))
const suffix = abbrev[order];
return (number / Math.pow(10, order * 3)).toFixed(precision) + suffix;
}
formatToUnits(12345, 2)
==> "12.35k"
formatToUnits(0, 3)
==> "0.000"
进一步改进Salman's Answer，因为像nFormatter(9999999,1)这样的情况返回1000K。
function formatNumberWithMetricPrefix(num, digits = 1) {
const si = [
{value: 1e18, symbol: 'E'},
{value: 1e15, symbol: 'P'},
{value: 1e12, symbol: 'T'},
{value: 1e9, symbol: 'G'},
{value: 1e6, symbol: 'M'},
{value: 1e3, symbol: 'k'},
{value: 0, symbol: ''},
];
const rx = /\.0+$|(\.[0-9]*[1-9])0+$/;
function divideNum(divider) {
return (num / (divider || 1)).toFixed(digits);
}
let i = si.findIndex(({value}) => num >= value);
if (+divideNum(si[i].value) >= 1e3 && si[i - 1]) {
i -= 1;
}
const {value, symbol} = si[i];
return divideNum(value).replace(rx, '$1') + symbol;
}
这里有一个简单的解决方案，可以避免所有的if语句(借助Math的力量)。
var SI_SYMBOL = ["", "k", "M", "G", "T", "P", "E"];
function abbreviateNumber(number){
// what tier? (determines SI symbol)
var tier = Math.log10(Math.abs(number)) / 3 | 0;
// if zero, we don't need a suffix
if(tier == 0) return number;
// get suffix and determine scale
var suffix = SI_SYMBOL[tier];
var scale = Math.pow(10, tier * 3);
// scale the number
var scaled = number / scale;
// format number and add suffix
return scaled.toFixed(1) + suffix;
}
奖金模因
SI代表什么?
我认为这是一个解决方案。
var unitlist = ["","K","M","G"]; function formatnumber(number){ let sign = Math.sign(number); let unit = 0; while(Math.abs(number) > 1000) { unit = unit + 1; number = Math.floor(Math.abs(number) / 100)/10; } console.log(sign*Math.abs(number) + unitlist[unit]); } formatnumber(999); formatnumber(1234); formatnumber(12345); formatnumber(123456); formatnumber(1234567); formatnumber(12345678); formatnumber(-999); formatnumber(-1234); formatnumber(-12345); formatnumber(-123456); formatnumber(-1234567); formatnumber(-12345678);
