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
进一步改进@Yash的回答,支持负数:
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"
其他回答
Waylon flynn的答案的修改版本,支持负指数:
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));
预期的输出:
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 }
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;
}
最简单最简单的方法是
new Intl.NumberFormat('en-IN', {
notation: "compact",
compactDisplay: "short",
style: 'currency',
currency: 'INR'
}).format(1000).replace("T", "K")
这对任何数字都适用。包括L、Cr等。
注意:不能在safari中工作。
这里有一个简单的解决方案,可以避免所有的if语句(借助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(Math.abs(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;
}
奖金模因
SI代表什么?
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);