我需要以等于一千,或1.1K,1.2K,1.9K等1K的格式显示货币值,如果它不是偶数千,否则如果低于一千,显示正常500,100,250等,使用javascript格式化数字?
听起来这应该适合你:
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
使用Math对象,映射对象,for循环,正则表达式等,这个线程上的很多答案变得相当复杂。但是这些方法并没有真正提高代码的可读性或性能。直接的方法似乎提供了最好的设计。
const formatCash = n => {
if (n < 1e3) return n;
if (n >= 1e3) return +(n / 1e3).toFixed(1) + "K";
};
console.log(formatCash(2500));
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));
这篇文章很老了,但我不知何故到达这篇文章寻找的东西。所以添加我的输入数字js现在是一天的一站式解决方案。它提供了大量方法来帮助格式化数字
添加最佳答案,这将给1000k而不是1.0k的1k
function kFormatter(num) {
return num > 999 ? num % 1000 === 0 ? (num/1000).toFixed(0) + 'k' : (num/1000).toFixed(1) + 'k' : num
}
!Number.isFinite
' K M G T P E Z Y'
,请将' K M'
更改为M
下面的代码是1K = 1024,如果你想要1K = 1000,将所有1024改为1000。
Number.prototype.prefix = function (precision = 2) {
var units = ' K M G T P E Z Y'.split(' ');
if (this < 0) {
return '-' + Math.abs(this).prefix(precision);
}
if (this < 1) {
return this + units[0];
}
var power = Math.min(
Math.floor(Math.log(this) / Math.log(1024)),
units.length - 1
);
return (this / Math.pow(1024, power)).toFixed(precision) + units[power];
}
console.log('10240 = ' + (10240).prefix()) // 10.00K
console.log('1234000 = ' + (1234000).prefix(1)) // 1.2M
console.log('10000 = ' + (-10000).prefix()) // -9.77K
/*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;
}
Waylon Flinn的修改版本的答案支持负面指数:
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));
预期产量:
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 }
此函数可以将大量数字(正面和负面)转换为读者友好格式,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);
我正在使用这个功能。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 + $suffix : 0;
}
我决定在@ Novellizator的答案上扩展很多,以满足我的需求。我想要一个灵活的功能来处理我的大多数格式需求,而无需外部库。
let x = 1234567.8;
formatNumber(x); // '1,234,568'
formatNumber(x, {useOrderSuffix: true}); // '1M'
formatNumber(x, {useOrderSuffix: true, decimals: 3, maxOrder: 1}); // '1,234.568k'
formatNumber(x, {decimals: 2, style: '$'}); // '$1,234,567.80'
x = 10.615;
formatNumber(x, {style: '%'}); // '1,062%'
formatNumber(x, {useOrderSuffix: true, decimals: 1, style: '%'}); // '1.1k%'
formatNumber(x, {useOrderSuffix: true, decimals: 5, style: '%', minOrder: 2}); // '0.00106M%'
formatNumber(-Infinity); // '-∞'
formatNumber(NaN); // ''
formatNumber(NaN, {valueIfNaN: NaN}); // NaN
/*
* Return the given number as a formatted string. The default format is a plain
* integer with thousands-separator commas. The optional parameters facilitate
* other formats:
* - decimals = the number of decimals places to round to and show
* - valueIfNaN = the value to show for non-numeric input
* - style
* - '%': multiplies by 100 and appends a percent symbol
* - '$': 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 === '%' ? '%' : '');
}
更通用的版本:
function nFormatter(num, digits) {
var si = [
{ 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" }
];
var rx = /\.0+$|(\.[0-9]*[1-9])0+$/;
var i;
for (i = si.length - 1; i > 0; i--) {
if (num >= si[i].value) {
break;
}
}
return (num / si[i].value).toFixed(digits).replace(rx, "$1") + si[i].symbol;
}
/*
* Tests
*/
var tests = [
{ num: 1234, digits: 1 },
{ num: 100000000, digits: 1 },
{ num: 299792458, digits: 1 },
{ num: 759878, digits: 1 },
{ num: 759878, digits: 0 },
{ num: 123, digits: 1 },
{ num: 123.456, digits: 1 },
{ num: 123.456, digits: 2 },
{ num: 123.456, digits: 4 }
];
var i;
for (i = 0; i < tests.length; i++) {
console.log("nFormatter(" + tests[i].num + ", " + tests[i].digits + ") = " + nFormatter(tests[i].num, tests[i].digits));
}
进一步改进Salman的答案,因为它将nFormatter(33000)返回为33.0K
function nFormatter(num) {
if (num >= 1000000000) {
return (num / 1000000000).toFixed(1).replace(/\.0$/, '') + '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;
}
现在nFormatter(33000)= 33K
这是一个简单的解决方案,避免所有if
语句(具有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(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;
}
/**
* 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;
}
感谢@Cos评论,我删除了Math.round10依赖项。
这是从他更优雅的处理负数和“.0”案例的方法中得到改善。
你拥有的循环和“如果”案例越少,IMO越好。
function abbreviateNumber(number) {
var SI_POSTFIXES = ["", "k", "M", "G", "T", "P", "E"];
var tier = Math.log10(Math.abs(number)) / 3 | 0;
if(tier == 0) return number;
var postfix = SI_POSTFIXES[tier];
var scale = Math.pow(10, tier * 3);
var scaled = number / scale;
var formatted = scaled.toFixed(1) + '';
if (/\.0$/.test(formatted))
formatted = formatted.substr(0, formatted.length - 2);
return formatted + postfix;
}
jsFiddle with test cases - > https://jsfiddle.net/xyug4nvz/7/
这是相当优雅的。
function formatToUnits(number, precision) {
const abbrev = ['', 'k', 'm', 'b', 't'];
const unrangifiedOrder = Math.floor(Math.log10(Math.abs(number)) / 3)
const order = Math.max(0, Math.min(unrangifiedOrder, abbrev.length -1 ))
const suffix = abbrev[order];
return (number / Math.pow(10, order * 3)).toFixed(precision) + suffix;
}
formatToUnits(12345, 2)
==> "12.35k"
formatToUnits(0, 3)
==> "0.000"
通过负数支持进一步改善@Yash的答案:
function nFormatter(num) {
isNegative = false
if (num < 0) {
isNegative = true
}
num = Math.abs(num)
if (num >= 1000000000) {
formattedNumber = (num / 1000000000).toFixed(1).replace(/\.0$/, '') + '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"