哇，这里有这么多答案。我想我会告诉你我是如何解决它的，因为它似乎是最容易阅读的，处理负数，并且在JavaScript的千位数范围内。它也很容易改变到你想要的或进一步扩展。

const symbols = [ { value: 1, symbol: '' }, { value: 1e3, symbol: 'k' }, { value: 1e6, symbol: 'M' }, { value: 1e9, symbol: 'G' }, { value: 1e12, symbol: 'T' }, { value: 1e15, symbol: 'P' }, { value: 1e18, symbol: 'E' } ]; function numberFormatter(num, digits) { const numToCheck = Math.abs(num); for (let i = symbols.length - 1; i >= 0; i--) { if (numToCheck >= symbols[i].value) { const newNumber = (num / symbols[i].value).toFixed(digits); return `${newNumber}${symbols[i].symbol}`; } } return '0'; } const tests = [ { num: 1234, digits: 1 }, { num: 100000000, digits: 1 }, { num: 299792458, digits: 1 }, { num: 759878, digits: 1 }, { num: -759878, digits: 0 }, { num: 123, digits: 1 }, { num: 123.456, digits: 1 }, { num: -123.456, digits: 2 }, { num: 123.456, digits: 4 } ]; for (let i = 0; i < tests.length; i++) { console.log(`numberFormatter(${tests[i].num}, ${tests[i].digits})=${numberFormatter(tests[i].num, tests[i].digits)}`); }

一个更普遍的版本:

function nFormatter(num, digits) { const lookup = [ { value: 1, symbol: "" }, { value: 1e3, symbol: "k" }, { value: 1e6, symbol: "M" }, { value: 1e9, symbol: "G" }, { value: 1e12, symbol: "T" }, { value: 1e15, symbol: "P" }, { value: 1e18, symbol: "E" } ]; const rx = /\.0+$|(\.[0-9]*[1-9])0+$/; var item = lookup.slice().reverse().find(function(item) { return num >= item.value; }); return item ? (num / item.value).toFixed(digits).replace(rx, "$1") + item.symbol : "0"; } /* * Tests */ const tests = [ { num: 0, digits: 1 }, { num: 12, digits: 1 }, { num: 1234, digits: 1 }, { num: 100000000, digits: 1 }, { num: 299792458, digits: 1 }, { num: 759878, digits: 1 }, { num: 759878, digits: 0 }, { num: 123, digits: 1 }, { num: 123.456, digits: 1 }, { num: 123.456, digits: 2 }, { num: 123.456, digits: 4 } ]; tests.forEach(function(test) { console.log("nFormatter(" + test.num + ", " + test.digits + ") = " + nFormatter(test.num, test.digits)); });

直接的方法具有最好的可读性，并且使用最少的内存。不需要过多地使用regex、map对象、Math对象、for-loops等。

使用K格式化现金值

const formatCash = n => { 如果(n < 1e3)返回n; if (n >= 1e3) return +(n / 1e3).toFixed(1) +“K”; }; console.log (formatCash (2500));

使用K M B T格式化现金值

const formatCash = n => { 如果(n < 1e3)返回n; 如果1 e3 & & n (n > = < 1 e6)返回+ (n / 1 e3) .toFixed(1) +“K”; 如果1 e6 & & n (n > = < 1 e9) + 1 (n / e6)返回.toFixed(1) +“M”; if (n >= 1e9 && n < 1e12) return +(n / 1e9).toFixed(1) + "B"; if (n >= 1e12) return +(n / 1e12).toFixed(1) + "T"; }; console.log (formatCash (1235000));

使用负数

let format;
const number = -1235000;

if (number < 0) {
  format = '-' + formatCash(-1 * number);
} else {
  format = formatCash(number);
}

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

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;
}

function AmountConveter(amount) {
  return Math.abs(amount) > 999
    ? Math.sign(amount) * (Math.abs(amount) / 1000).toFixed(1) + "k"
    : Math.sign(amount) * Math.abs(amount);
}

console.log(AmountConveter(1200)); // 1.2k
console.log(AmountConveter(-1200)); // -1.2k
console.log(AmountConveter(900)); // 900
console.log(AmountConveter(-900)); // -900