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;
}

听起来这应该对你有用:

函数 kFormatter（num） { 返回 Math.abs（num） > 999 ？Math.sign（num）*（（Math.abs（num）/1000）.toFixed（1）） + 'k' ： Math.sign（num）*Math.abs（num） } console.log（kFormatter（1200））;1.2k console.log（kFormatter（-1200））;-1.2k console.log（kFormatter（900））;900 console.log（kFormatter（-900））;-900

最多支持数量。MAX_SAFE_INTEGER到Number。MIN_SAFE_INTEGER

function abbreviateThousands(value) { const num = Number(value) const absNum = Math.abs(num) const sign = Math.sign(num) const numLength = Math.round(absNum).toString().length const symbol = ['K', 'M', 'B', 'T', 'Q'] const symbolIndex = Math.floor((numLength - 1) / 3) - 1 const abbrv = symbol[symbolIndex] || symbol[symbol.length - 1] let divisor = 0 if (numLength > 15) divisor = 1e15 else if (numLength > 12) divisor = 1e12 else if (numLength > 9) divisor = 1e9 else if (numLength > 6) divisor = 1e6 else if (numLength > 3) divisor = 1e3 else return num return `${((sign * absNum) / divisor).toFixed(divisor && 1)}${abbrv}` } console.log(abbreviateThousands(234523452345)) // 234.5b (billion) console.log(abbreviateThousands(Number.MIN_SAFE_INTEGER)) // -9.0q (quadrillion)

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?

简单通用的方法

可以将COUNT_FORMATS配置对象设置为您想要的长度或长度，这取决于您测试的值范围。

// Configuration const COUNT_FORMATS = [ { // 0 - 999 letter: '', limit: 1e3 }, { // 1,000 - 999,999 letter: 'K', limit: 1e6 }, { // 1,000,000 - 999,999,999 letter: 'M', limit: 1e9 }, { // 1,000,000,000 - 999,999,999,999 letter: 'B', limit: 1e12 }, { // 1,000,000,000,000 - 999,999,999,999,999 letter: 'T', limit: 1e15 } ]; // Format Method: function formatCount(value) { const format = COUNT_FORMATS.find(format => (value < format.limit)); value = (1000 * value / format.limit); value = Math.round(value * 10) / 10; // keep one decimal number, only if needed return (value + format.letter); } // Test: const test = [274, 1683, 56512, 523491, 9523489, 5729532709, 9421032489032]; test.forEach(value => console.log(`${ value } >>> ${ formatCount(value) }`));

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