听起来这应该对你有用:

function kFormatter(num) {
return 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

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));
});

进一步改进Salman's Answer，因为它将nFormatter(33000)返回为33.0K

function nFormatter(num) {
if (num >= 1000000000) {
return (num / 1000000000).toFixed(1).replace(/\.0$/, '') + 'G';
}
if (num >= 1000000) {
return (num / 1000000).toFixed(1).replace(/\.0$/, '') + 'M';
}
if (num >= 1000) {
return (num / 1000).toFixed(1).replace(/\.0$/, '') + 'K';
}
return num;
}

now nFormatter(33000) = 33K

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"

/**
* Shorten number to thousands, millions, billions, etc.
* http://en.wikipedia.org/wiki/Metric_prefix
*
* @param {number} num Number to shorten.
* @param {number} [digits=0] The number of digits to appear after the decimal point.
* @returns {string|number}
*
* @example
* // returns '12.5k'
* shortenLargeNumber(12543, 1)
*
* @example
* // returns '-13k'
* shortenLargeNumber(-12567)
*
* @example
* // returns '51M'
* shortenLargeNumber(51000000)
*
* @example
* // returns 651
* shortenLargeNumber(651)
*
* @example
* // returns 0.12345
* shortenLargeNumber(0.12345)
*/
function shortenLargeNumber(num, digits) {
var units = ['k', 'M', 'G', 'T', 'P', 'E', 'Z', 'Y'],
decimal;


for(var i=units.length-1; i>=0; i--) {
decimal = Math.pow(1000, i+1);


if(num <= -decimal || num >= decimal) {
return +(num / decimal).toFixed(digits) + units[i];
}
}


return num;
}

谢谢@Cos的评论，我删除了Math。round10依赖。

这篇文章很旧了，但我不知何故找到了这篇文章。所以添加我的输入数字js是一站式的解决方案现在一天。它提供了大量的方法来帮助格式化数字

http://numeraljs.com/

你可以使用模仿Python高级字符串格式化PEP3101的d3-format包:

var f = require('d3-format')
console.log(f.format('.2s')(2500)) // displays "2.5k"

这里有一个简单的解决方案，可以避免所有if语句(使用Math的功能)。

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;
}

奖金文化基因

SI代表什么?< / >

如果你喜欢，就把功劳归于韦伦·弗林

这比他处理负数和“。0"的情况。

更少的循环和“if”;在我看来，你的案子越多越好。

function abbreviateNumber(number) {
const SI_POSTFIXES = ["", "k", "M", "G", "T", "P", "E"];
const sign = number < 0 ? '-1' : '';
const absNumber = Math.abs(number);
const tier = Math.log10(absNumber) / 3 | 0;
// if zero, we don't need a prefix
if(tier == 0) return `${absNumber}`;
// get postfix and determine scale
const postfix = SI_POSTFIXES[tier];
const scale = Math.pow(10, tier * 3);
// scale the number
const scaled = absNumber / scale;
const floored = Math.floor(scaled * 10) / 10;
// format number and add postfix as suffix
let str = floored.toFixed(1);
// remove '.0' case
str = (/\.0$/.test(str)) ? str.substr(0, str.length - 2) : str;
return `${sign}${str}${postfix}`;
}

jsFiddle测试用例->https://jsfiddle.net/qhbrz04o/9/

加上上面的答案，这将给出1000的1k而不是1.0k

function kFormatter(num) {
return num > 999 ? num % 1000 === 0 ? (num/1000).toFixed(0) + 'k' : (num/1000).toFixed(1) + 'k' : num
}

/*including negative values*/
function nFormatter(num) {
let neg = false;
if(num < 0){
num = num * -1;
neg = true;
}
if (num >= 1000000000) {
if(neg){
return -1 * (num / 1000000000).toFixed(1).replace(/\.0$/, '') + 'G';
}
return (num / 1000000000).toFixed(1).replace(/\.0$/, '') + 'G';
}
if (num >= 1000000) {
if(neg){
return -1 * (num / 1000000).toFixed(1).replace(/\.0$/, '') + 'M';
}
return (num / 1000000).toFixed(1).replace(/\.0$/, '') + 'M';
}
if (num >= 1000) {
if(neg){
return -1 * (num / 1000).toFixed(1).replace(/\.0$/, '') + 'K';
}
return (num / 1000).toFixed(1).replace(/\.0$/, '') + 'K';
}
return num;
}

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"

• 支持负数
• 检查!isFinite
• 如果你想要最大单位是M，将' K M G T P E Z Y'更改为' K M'
• 基数选项(1K = 1000 / 1K = 1024)

Number.prototype.prefix = function (precision, base) {


var units = ' K M G T P E Z Y'.split(' ');


if (typeof precision === 'undefined') {
precision = 2;
}


if (typeof base === 'undefined') {
base = 1000;
}


if (this == 0 || !isFinite(this)) {
return this.toFixed(precision) + units[0];
}


var power = Math.floor(Math.log(Math.abs(this)) / Math.log(base));
// Make sure not larger than max prefix
power = Math.min(power, units.length - 1);


return (this / Math.pow(base, power)).toFixed(precision) + units[power];
};


console.log('0 = ' + (0).prefix()) // 0.00
console.log('10000 = ' + (10000).prefix()) // 10.00K
console.log('1234000 = ' + (1234000).prefix(1)) // 1.2M
console.log('-10000 = ' + (-10240).prefix(1, 1024)) // -10.0K
console.log('-Infinity = ' + (-Infinity).prefix()) // -Infinity
console.log('NaN = ' + (NaN).prefix()) // NaN

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 }

这个函数可以转换巨大的数字(正数和amp;Negative)转换成一种读者友好的格式，同时又不失其准确性:

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);

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 => {
if (n < 1e3) return 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 => {
if (n < 1e3) return n;
if (n >= 1e3 && n < 1e6) return +(n / 1e3).toFixed(1) + "K";
if (n >= 1e6 && n < 1e9) return +(n / 1e6).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);
}

我用的是这个函数。它适用于php和javascript。

    /**
* @param $n
* @return string
* Use to convert large positive numbers in to short form like 1K+, 100K+, 199K+, 1M+, 10M+, 1B+ etc
*/
function num_format($n) {
$n_format = null;
$suffix = null;
if ($n > 0 && $n < 1000) {
$n_format = Math.floor($n);
$suffix = '';
}
else if ($n == 1000) {
$n_format = Math.floor($n / 1000);   //For PHP only use floor function insted of Math.floor()
$suffix = 'K';
}
else if ($n > 1000 && $n < 1000000) {
$n_format = Math.floor($n / 1000);
$suffix = 'K+';
} else if ($n == 1000000) {
$n_format = Math.floor($n / 1000000);
$suffix = 'M';
} else if ($n > 1000000 && $n < 1000000000) {
$n_format = Math.floor($n / 1000000);
$suffix = 'M+';
} else if ($n == 1000000000) {
$n_format = Math.floor($n / 1000000000);
$suffix = 'B';
} else if ($n > 1000000000 && $n < 1000000000000) {
$n_format = Math.floor($n / 1000000000);
$suffix = 'B+';
} else if ($n == 1000000000000) {
$n_format = Math.floor($n / 1000000000000);
$suffix = 'T';
} else if ($n >= 1000000000000) {
$n_format = Math.floor($n / 1000000000000);
$suffix = 'T+';
}




/***** For PHP  ******/
//  return !empty($n_format . $suffix) ? $n_format . $suffix : 0;


/***** For Javascript ******/
return ($n_format + $suffix).length > 0 ? $n_format + $suffix : 0;
}

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
  李< / ul > < / >
  • 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 === '%' ? '%' : '');
  }
  

改进@tfmontague的答案，进一步格式化小数点。33.0k到33k

largeNumberFormatter(value: number): any {
let result: any = value;


if (value >= 1e3 && value < 1e6) { result = (value / 1e3).toFixed(1).replace(/\.0$/, '') + 'K'; }
if (value >= 1e6 && value < 1e9) { result = (value / 1e6).toFixed(1).replace(/\.0$/, '') + 'M'; }
if (value >= 1e9) { result = (value / 1e9).toFixed(1).replace(/\.0$/, '') + 'T'; }


return result;
}

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)}`);
}

不满足任何张贴的解决方案，所以这是我的版本:

1. 支持正负两位数
2. 支持负指数
3. 如果可能，四舍五入到下一个指数
4. 执行边界检查(对于非常大/小的数字不会出错)
5. 去掉后面的零/空格

6. 支持精度参数

   function abbreviateNumber(number,digits=2) {
   var expK = Math.floor(Math.log10(Math.abs(number)) / 3);
   var scaled = number / Math.pow(1000, expK);
   
   
   if(Math.abs(scaled.toFixed(digits))>=1000) { // Check for rounding to next exponent
   scaled /= 1000;
   expK += 1;
   }
   
   
   var SI_SYMBOLS = "apμm kMGTPE";
   var BASE0_OFFSET = SI_SYMBOLS.indexOf(' ');
   
   
   if (expK + BASE0_OFFSET>=SI_SYMBOLS.length) { // Bound check
   expK = SI_SYMBOLS.length-1 - BASE0_OFFSET;
   scaled = number / Math.pow(1000, expK);
   }
   else if (expK + BASE0_OFFSET < 0) return 0;  // Too small
   
   
   return scaled.toFixed(digits).replace(/(\.|(\..*?))0+$/,'$2') + SI_SYMBOLS[expK+BASE0_OFFSET].trim();
   }
   
   
   //////////////////
   
   
   const tests = [
   [0.0000000000001,2],
   [0.00000000001,2],
   [0.000000001,2],
   [0.000001,2],
   [0.001,2],
   [0.0016,2],
   [-0.0016,2],
   [0.01,2],
   [1,2],
   [999.99,2],
   [999.99,1],
   [-999.99,1],
   [999999,2],
   [999999999999,2],
   [999999999999999999,2],
   [99999999999999999999,2],
   ];
   
   
   for (var i = 0; i < tests.length; i++) {
   console.log(abbreviateNumber(tests[i][0], tests[i][1]) );
   }

   李< / p > < / >

我想出了一个非常编码的高尔夫球，它非常短!

var beautify=n=>((Math.log10(n)/3|0)==0)?n:Number((n/Math.pow(10,(Math.log10(n)/3|0)*3)).toFixed(1))+["","K","M","B","T",][Math.log10(n)/3|0];


console.log(beautify(1000))
console.log(beautify(10000000))

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;
}

简单通用的方法

你可以将COUNT_FORMATS配置对象设置为你想要的长度或长度，这取决于你测试的值范围。

// 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) }`));

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。

ES2020在Intl.NumberFormat中增加了对此的支持，使用的符号如下:

let formatter = Intl.NumberFormat('en', { notation: 'compact' });
// example 1
let million = formatter.format(1e6);
// example 2
let billion = formatter.format(1e9);
// print
console.log(million == '1M', billion == '1B');

• https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Intl/NumberFormat/NumberFormat
• https://tc39.es/ecma402#numberformat-objects

通过消除@martin-sznapka解决方案中的循环，您将减少40%的执行时间。

function formatNum(num,digits) {
let units = ['k', 'M', 'G', 'T', 'P', 'E', 'Z', 'Y'];
let floor = Math.floor(Math.abs(num).toString().length / 3);
let value=+(num / Math.pow(1000, floor))
return value.toFixed(value > 1?digits:2) + units[floor - 1];


}

速度测试(200000随机样本)从这个线程不同的解决方案

Execution time: formatNum          418  ms
Execution time: kFormatter         438  ms it just use "k" no "M".."T"
Execution time: beautify           593  ms doesnt support - negatives
Execution time: shortenLargeNumber 682  ms
Execution time: Intl.NumberFormat  13197ms

一个简短的替代方案:

function nFormatter(num) {
const format = [
{ 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: 1, symbol: '' },
];
const formatIndex = format.findIndex((data) => num >= data.value);
console.log(formatIndex)
return (num / format[formatIndex === -1? 6: formatIndex].value).toFixed(2) + format[formatIndex === -1?6: formatIndex].symbol;
}

支持上至Number.MAX_SAFE_INTEGER，下至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)

以下是我对韦伦·弗林的回答的看法。这将删除.0并修复当层不是整数时的未定义。

const SI_SYMBOL = ['', 'k', 'M', 'G', 'T', 'P', 'E'];


abbreviateNumber(num) {
const tier = Math.floor(Math.log10(num) / 3) || 0;
let result = '' + num;
// if zero, we don't need a suffix
if (tier > 0) {
// get suffix and determine scale
const suffix = SI_SYMBOL[tier];
const scale = Math.pow(10, tier * 3);
// scale the number
const scaled = num / scale;
// format number and add suffix
result = scaled.toFixed(1).replace('.0', '') + suffix;
}
return result;
}

下面是一个使用for的选项:

function numberFormat(d) {
for (var e = 0; d >= 1000; e++) {
d /= 1000;
}
return d.toFixed(3) + ['', ' k', ' M', ' G'][e];
}


let s = numberFormat(9012345678);
console.log(s == '9.012 G');

韦伦·弗林的解决方案的2020版本。

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;
};

测试和示例:

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)}`);
});
});

你可以用最简单的方法。

$('#attrib-id').val(Number(response.column/1000000).toLocaleString()); // Million

您可以将值除以您想要的数字来使用其他属性，例如如果您想显示"K"在一个数字前面，你应该使用Number(response.column/1000)，以及其他相应的东西。

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);

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"

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