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
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;
}
其他回答
这个函数可以将巨大的数字(正数和负数)转换为读者友好的格式，而不会失去其精度:
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);
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 === '%' ? '%' : '');
}
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/
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 => { 如果(n < 1e3)返回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 => { 如果(n < 1e3)返回n; 如果1 e3 & & n (n > = < 1 e6)返回+ (n / 1 e3) .toFixed(1) +“K”; 如果1 e6 & & n (n > = < 1 e9) + 1 (n / e6)返回.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);
}
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)}`); }); });
