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
您可以使用模仿Python高级字符串格式化PEP3101的d3格式包:
var f = require('d3-format')
console.log(f.format('.2s')(2500)) // displays "2.5k"
其他回答
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}${postfix}`;
}
jsFiddle测试用例-> https://jsfiddle.net/qhbrz04o/9/
您可以使用模仿Python高级字符串格式化PEP3101的d3格式包:
var f = require('d3-format')
console.log(f.format('.2s')(2500)) // displays "2.5k"
这里有一个简单的解决方案,可以避免所有的if语句(借助Math的力量)。
var SI_SYMBOL = ["", "k", "M", "G", "T", "P", "E"];
function abbreviateNumber(number){
// what tier? (determines SI symbol)
var tier = Math.log10(Math.abs(number)) / 3 | 0;
// if zero, we don't need a suffix
if(tier == 0) return number;
// get suffix and determine scale
var suffix = SI_SYMBOL[tier];
var scale = Math.pow(10, tier * 3);
// scale the number
var scaled = number / scale;
// format number and add suffix
return scaled.toFixed(1) + suffix;
}
奖金模因
SI代表什么?
加上上面的答案,这将给出1000的1k而不是1.0k
function kFormatter(num) {
return num > 999 ? num % 1000 === 0 ? (num/1000).toFixed(0) + 'k' : (num/1000).toFixed(1) + 'k' : num
}
进一步改进Salman's Answer,因为像nFormatter(9999999,1)这样的情况返回1000K。
function formatNumberWithMetricPrefix(num, digits = 1) {
const si = [
{value: 1e18, symbol: 'E'},
{value: 1e15, symbol: 'P'},
{value: 1e12, symbol: 'T'},
{value: 1e9, symbol: 'G'},
{value: 1e6, symbol: 'M'},
{value: 1e3, symbol: 'k'},
{value: 0, symbol: ''},
];
const rx = /\.0+$|(\.[0-9]*[1-9])0+$/;
function divideNum(divider) {
return (num / (divider || 1)).toFixed(digits);
}
let i = si.findIndex(({value}) => num >= value);
if (+divideNum(si[i].value) >= 1e3 && si[i - 1]) {
i -= 1;
}
const {value, symbol} = si[i];
return divideNum(value).replace(rx, '$1') + symbol;
}