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

听起来这应该对你有用:

函数 kFormatter(num) { 返回 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

其他回答

简单通用的方法

可以将COUNT_FORMATS配置对象设置为您想要的长度或长度,这取决于您测试的值范围。

// Configuration const COUNT_FORMATS = [ { // 0 - 999 letter: '', limit: 1e3 }, { // 1,000 - 999,999 letter: 'K', limit: 1e6 }, { // 1,000,000 - 999,999,999 letter: 'M', limit: 1e9 }, { // 1,000,000,000 - 999,999,999,999 letter: 'B', limit: 1e12 }, { // 1,000,000,000,000 - 999,999,999,999,999 letter: 'T', limit: 1e15 } ]; // Format Method: function formatCount(value) { const format = COUNT_FORMATS.find(format => (value < format.limit)); value = (1000 * value / format.limit); value = Math.round(value * 10) / 10; // keep one decimal number, only if needed return (value + format.letter); } // Test: const test = [274, 1683, 56512, 523491, 9523489, 5729532709, 9421032489032]; test.forEach(value => console.log(`${ value } >>> ${ formatCount(value) }`));

下面是一个使用for的选项:

函数numberFormat(d) For (var e = 0;D >= 1000;e + +) { D /= 1000; } return d.toFixed(3) + [", ' k', ' M', ' G'][e]; } let s = numberFormat(9012345678); console.log(s == '9.012 G');

听起来这应该对你有用:

函数 kFormatter(num) { 返回 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

我认为这是一个解决方案。

var unitlist = ["","K","M","G"]; function formatnumber(number){ let sign = Math.sign(number); let unit = 0; while(Math.abs(number) > 1000) { unit = unit + 1; number = Math.floor(Math.abs(number) / 100)/10; } console.log(sign*Math.abs(number) + unitlist[unit]); } formatnumber(999); formatnumber(1234); formatnumber(12345); formatnumber(123456); formatnumber(1234567); formatnumber(12345678); formatnumber(-999); formatnumber(-1234); formatnumber(-12345); formatnumber(-123456); formatnumber(-1234567); formatnumber(-12345678);

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/