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

韦伦·弗林解决方案的2020版。

const SI_SYMBOLS = ["", "k", "M", "G", "T", "P", "E"];

const abbreviateNumber = (number, minDigits, maxDigits) => {
    if (number === 0) return number;

    // determines SI symbol
    const tier = Math.floor(Math.log10(Math.abs(number)) / 3);

    // get suffix and determine scale
    const suffix = SI_SYMBOLS[tier];
    const scale = 10 ** (tier * 3);

    // scale the number
    const scaled = number / scale;

    // format number and add suffix
    return scaled.toLocaleString(undefined, {
        minimumFractionDigits: minDigits,
        maximumFractionDigits: maxDigits,
    }) + suffix;
};

Tests and examples: const abbreviateNumberFactory = (symbols) => ( (number, minDigits, maxDigits) => { if (number === 0) return number; // determines SI symbol const tier = Math.floor(Math.log10(Math.abs(number)) / 3); // get suffix and determine scale const suffix = symbols[tier]; const scale = 10 ** (tier * 3); // scale the number const scaled = number / scale; // format number and add suffix return scaled.toLocaleString(undefined, { minimumFractionDigits: minDigits, maximumFractionDigits: maxDigits, }) + suffix; } ); const SI_SYMBOLS = ["", "k", "M", "G", "T", "P", "E"]; const SHORT_SYMBOLS = ["", "K", "M", "B", "T", "Q"]; const LONG_SYMBOLS = ["", " thousand", " million", " billion", " trillion", " quadrillion"]; const abbreviateNumberSI = abbreviateNumberFactory(SI_SYMBOLS); const abbreviateNumberShort = abbreviateNumberFactory(SHORT_SYMBOLS); const abbreviateNumberLong = abbreviateNumberFactory(LONG_SYMBOLS); const tests = [1e5, -9e7, [1009999.999, 2], [245345235.34513, 1, 1], [-72773144123, 3] ]; const functions = { abbreviateNumberSI, abbreviateNumberShort, abbreviateNumberLong, }; tests.forEach((test) => { const testValue = Array.isArray(test) ? test : [test]; Object.entries(functions).forEach(([key, func]) => { console.log(`${key}(${testValue.join(', ')}) = ${func(...testValue)}`); }); });

其他回答

function   transform(value,args) {
    const suffixes = ['K', 'M', 'B', 'T', 'P', 'E'];

    if (!value) {
      return null;
    }

    if (Number.isNaN(value)) {
      return null;
    }

    if (value < 1000) {
      return value;
    }

    const exp = Math.floor(Math.log(value) / Math.log(1000));

    const returnValue = (value / Math.pow(1000, exp)).toFixed(args) + suffixes[exp - 1];

    return returnValue;
  }

变换(9999 2)

/ /“10.00 k”

进一步改进@Yash的回答,支持负数:

function nFormatter(num) {
    isNegative = false
    if (num < 0) {
        isNegative = true
    }
    num = Math.abs(num)
    if (num >= 1000000000) {
        formattedNumber = (num / 1000000000).toFixed(1).replace(/\.0$/, '') + 'G';
    } else if (num >= 1000000) {
        formattedNumber =  (num / 1000000).toFixed(1).replace(/\.0$/, '') + 'M';
    } else  if (num >= 1000) {
        formattedNumber =  (num / 1000).toFixed(1).replace(/\.0$/, '') + 'K';
    } else {
        formattedNumber = num;
    }   
    if(isNegative) { formattedNumber = '-' + formattedNumber }
    return formattedNumber;
}

nFormatter(-120000)
"-120K"
nFormatter(120000)
"120K"

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?

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"

Waylon flynn的答案的修改版本,支持负指数:

function metric(number) { const SI_SYMBOL = [ ["", "k", "M", "G", "T", "P", "E"], // + ["", "m", "μ", "n", "p", "f", "a"] // - ]; const tier = Math.floor(Math.log10(Math.abs(number)) / 3) | 0; const n = tier < 0 ? 1 : 0; const t = Math.abs(tier); const scale = Math.pow(10, tier * 3); return { number: number, symbol: SI_SYMBOL[n][t], scale: scale, scaled: number / scale } } function metric_suffix(number, precision) { const m = metric(number); return (typeof precision === 'number' ? m.scaled.toFixed(precision) : m.scaled) + m.symbol; } for (var i = 1e-6, s = 1; i < 1e7; i *= 10, s *= -1) { // toggles sign in each iteration console.log(metric_suffix(s * (i + i / 5), 1)); } console.log(metric(0));

预期的输出:

   1.2μ
 -12.0μ
 120.0μ
  -1.2m
  12.0m
-120.0m
   1.2
 -12.0
 120.0
  -1.2k
  12.0k
-120.0k
   1.2M
{ number: 0, symbol: '', scale: 1, scaled: 0 }