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);
其他回答
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
}
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; }
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。
循环和“如果”情况越少，IMO就越好。
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/
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;
}
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;
}
