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
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 + $suffix : 0;
}
其他回答
/*including negative values*/
function nFormatter(num) {
let neg = false;
if(num < 0){
num = num * -1;
neg = true;
}
if (num >= 1000000000) {
if(neg){
return -1 * (num / 1000000000).toFixed(1).replace(/\.0$/, '') + '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$/, '') + 'K';
}
return num;
}
这个函数可以将巨大的数字(正数和负数)转换为读者友好的格式，而不会失去其精度:
function abbrNum(n) { if (!n || (n && typeof n !== 'number')) { return ''; } const ranges = [ { divider: 1e12 , suffix: 't' }, { divider: 1e9 , suffix: 'b' }, { divider: 1e6 , suffix: 'm' }, { divider: 1e3 , suffix: 'k' } ]; const range = ranges.find(r => Math.abs(n) >= r.divider); if (range) { return (n / range.divider).toString() + range.suffix; } return n.toString(); } /* test cases */ let testAry = [99, 1200, -150000, 9000000]; let resultAry = testAry.map(abbrNum); console.log("result array: " + resultAry);
加上上面的答案，这将给出1000的1k而不是1.0k
function kFormatter(num) {
return num > 999 ? num % 1000 === 0 ? (num/1000).toFixed(0) + 'k' : (num/1000).toFixed(1) + 'k' : num
}
我决定在这里扩展@Novellizator的答案，以满足我的需求。我想要一个灵活的函数来处理我的大部分格式化需求，而不需要外部库。
特性
选择使用顺序后缀(k, M等) 选项指定要使用的订单后缀的自定义列表 选项来约束最小和最大顺序 控制小数点后的位数 自动顺序分隔逗号 可选百分比或美元格式 控制在非数字输入的情况下返回什么 适用于负数和无穷数
例子
let x = 1234567.8;
formatNumber(x); // '1,234,568'
formatNumber(x, {useOrderSuffix: true}); // '1M'
formatNumber(x, {useOrderSuffix: true, decimals: 3, maxOrder: 1}); // '1,234.568k'
formatNumber(x, {decimals: 2, style: '$'}); // '$1,234,567.80'
x = 10.615;
formatNumber(x, {style: '%'}); // '1,062%'
formatNumber(x, {useOrderSuffix: true, decimals: 1, style: '%'}); // '1.1k%'
formatNumber(x, {useOrderSuffix: true, decimals: 5, style: '%', minOrder: 2}); // '0.00106M%'
formatNumber(-Infinity); // '-∞'
formatNumber(NaN); // ''
formatNumber(NaN, {valueIfNaN: NaN}); // NaN
函数
/*
* Return the given number as a formatted string. The default format is a plain
* integer with thousands-separator commas. The optional parameters facilitate
* other formats:
* - decimals = the number of decimals places to round to and show
* - valueIfNaN = the value to show for non-numeric input
* - style
* - '%': multiplies by 100 and appends a percent symbol
* - '$': 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 === '%' ? '%' : '');
}
ES2020在Intl中增加了对此的支持。使用如下表示法:
let formatter = Intl。NumberFormat('en'，{符号:'紧凑'}); //示例1 让million = formatter.format(1e6); //示例2 Let billion = formatter.format(1e9); / /打印 console.log(million == '1M'， billion == '1B');
注意如上所示，第二个示例生成1B而不是1G。 NumberFormat规格:
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Intl/NumberFormat/NumberFormat https://tc39.es/ecma402#numberformat-objects
注意，目前并不是所有的浏览器都支持ES2020，所以你可能需要这个 Polyfill: https://formatjs.io/docs/polyfills/intl-numberformat
