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

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)}`); }); });

我决定在这里扩展@Novellizator的答案,以满足我的需求。我想要一个灵活的函数来处理我的大部分格式化需求,而不需要外部库。

特性

选择使用顺序后缀(k, M等) 选项指定要使用的订单后缀的自定义列表 选项来约束最小和最大顺序 控制小数点后的位数 自动顺序分隔逗号 可选百分比或美元格式 控制在非数字输入的情况下返回什么 适用于负数和无穷数

例子

let x = 1234567.8;
formatNumber(x);  // '1,234,568'
formatNumber(x, {useOrderSuffix: true});  // '1M'
formatNumber(x, {useOrderSuffix: true, decimals: 3, maxOrder: 1});  // '1,234.568k'
formatNumber(x, {decimals: 2, style: '$'});  // '$1,234,567.80'

x = 10.615;
formatNumber(x, {style: '%'});  // '1,062%'
formatNumber(x, {useOrderSuffix: true, decimals: 1, style: '%'});  // '1.1k%'
formatNumber(x, {useOrderSuffix: true, decimals: 5, style: '%', minOrder: 2});  // '0.00106M%'

formatNumber(-Infinity);  // '-∞'
formatNumber(NaN);  // ''
formatNumber(NaN, {valueIfNaN: NaN});  // NaN

函数

/*
 * Return the given number as a formatted string.  The default format is a plain
 * integer with thousands-separator commas.  The optional parameters facilitate
 * other formats:
 *   - decimals = the number of decimals places to round to and show
 *   - valueIfNaN = the value to show for non-numeric input
 *   - style
 *     - '%': multiplies by 100 and appends a percent symbol
 *     - '$': prepends a dollar sign
 *   - useOrderSuffix = whether to use suffixes like k for 1,000, etc.
 *   - orderSuffixes = the list of suffixes to use
 *   - minOrder and maxOrder allow the order to be constrained.  Examples:
 *     - minOrder = 1 means the k suffix should be used for numbers < 1,000
 *     - maxOrder = 1 means the k suffix should be used for numbers >= 1,000,000
 */
function formatNumber(number, {
    decimals = 0,
    valueIfNaN = '',
    style = '',
    useOrderSuffix = false,
    orderSuffixes = ['', 'k', 'M', 'B', 'T'],
    minOrder = 0,
    maxOrder = Infinity
  } = {}) {

  let x = parseFloat(number);

  if (isNaN(x))
    return valueIfNaN;

  if (style === '%')
    x *= 100.0;

  let order;
  if (!isFinite(x) || !useOrderSuffix)
    order = 0;
  else if (minOrder === maxOrder)
    order = minOrder;
  else {
    const unboundedOrder = Math.floor(Math.log10(Math.abs(x)) / 3);
    order = Math.max(
      0,
      minOrder,
      Math.min(unboundedOrder, maxOrder, orderSuffixes.length - 1)
    );
  }

  const orderSuffix = orderSuffixes[order];
  if (order !== 0)
    x /= Math.pow(10, order * 3);

  return (style === '$' ? '$' : '') +
    x.toLocaleString(
      'en-US',
      {
        style: 'decimal',
        minimumFractionDigits: decimals,
        maximumFractionDigits: decimals
      }
    ) +
    orderSuffix +
    (style === '%' ? '%' : '');
}
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

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 }

最简单最简单的方法是

new Intl.NumberFormat('en-IN', { 
    notation: "compact",
    compactDisplay: "short",
    style: 'currency',
    currency: 'INR'
}).format(1000).replace("T", "K")

这对任何数字都适用。包括L、Cr等。

注意:不能在safari中工作。