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
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"
其他回答
/**
* 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。
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)}`); }
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;
}
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 } >>> ${ formatCount(value) }`));
我决定在这里扩展@Novellizator的答案,以满足我的需求。我想要一个灵活的函数来处理我的大部分格式化需求,而不需要外部库。
特性
选择使用顺序后缀(k, M等) 选项指定要使用的订单后缀的自定义列表 选项来约束最小和最大顺序 控制小数点后的位数 自动顺序分隔逗号 可选百分比或美元格式 控制在非数字输入的情况下返回什么 适用于负数和无穷数
例子
let x = 1234567.8;
formatNumber(x); // '1,234,568'
formatNumber(x, {useOrderSuffix: true}); // '1M'
formatNumber(x, {useOrderSuffix: true, decimals: 3, maxOrder: 1}); // '1,234.568k'
formatNumber(x, {decimals: 2, style: '$'}); // '$1,234,567.80'
x = 10.615;
formatNumber(x, {style: '%'}); // '1,062%'
formatNumber(x, {useOrderSuffix: true, decimals: 1, style: '%'}); // '1.1k%'
formatNumber(x, {useOrderSuffix: true, decimals: 5, style: '%', minOrder: 2}); // '0.00106M%'
formatNumber(-Infinity); // '-∞'
formatNumber(NaN); // ''
formatNumber(NaN, {valueIfNaN: NaN}); // NaN
函数
/*
* Return the given number as a formatted string. The default format is a plain
* integer with thousands-separator commas. The optional parameters facilitate
* other formats:
* - decimals = the number of decimals places to round to and show
* - valueIfNaN = the value to show for non-numeric input
* - style
* - '%': multiplies by 100 and appends a percent symbol
* - '$': 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 === '%' ? '%' : '');
}