如果数字是1000或更多，如何将数字格式化为2.5K，否则在javascript中是900 ?

I need to show a currency value in the format of 1K of equal to one thousand, or 1.1K, 1.2K, 1.9K etc, if its not an even thousands, otherwise if under a thousand, display normal 500, 100, 250 etc, using javascript to format the number?

我需要以1K = 1000，或1.1K, 1.2K, 1.9K等的格式显示货币值，如果不是上千，否则在1000以下，显示正常500,100,250等等，用javascript来格式化数字?

12 个解决方案

#1

Sounds like this should work for you:

听起来这对你应该有用:

function kFormatter(num) {
    return num > 999 ? (num/1000).toFixed(1) + 'k' : num
}

console.log(kFormatter(1200));
console.log(kFormatter(900));

#2

A more generalized version:

一个更通用的版本:

function nFormatter(num, digits) {
  var si = [
    { value: 1, symbol: "" },
    { value: 1E3, symbol: "k" },
    { value: 1E6, symbol: "M" },
    { value: 1E9, symbol: "G" },
    { value: 1E12, symbol: "T" },
    { value: 1E15, symbol: "P" },
    { value: 1E18, symbol: "E" }
  ];
  var rx = /\.0+$|(\.[0-9]*[1-9])0+$/;
  var i;
  for (i = si.length - 1; i > 0; i--) {
    if (num >= si[i].value) {
      break;
    }
  }
  return (num / si[i].value).toFixed(digits).replace(rx, "$1") + si[i].symbol;
}

/*
 * Tests
 */
var tests = [
  { num: 1234, digits: 1 },
  { num: 100000000, digits: 1 },
  { num: 299792458, digits: 1 },
  { num: 759878, digits: 1 },
  { num: 759878, digits: 0 },
  { num: 123, digits: 1 },
  { num: 123.456, digits: 1 },
  { num: 123.456, digits: 2 },
  { num: 123.456, digits: 4 }
];
var i;
for (i = 0; i < tests.length; i++) {
  console.log("nFormatter(" + tests[i].num + ", " + tests[i].digits + ") = " + nFormatter(tests[i].num, tests[i].digits));
}

#3

Further improving Salman's Answer because it returns nFormatter(33000) as 33.0K

进一步改进Salman的答案，因为它返回的nFormatter(33000)为33.0K

function nFormatter(num) {
     if (num >= 1000000000) {
        return (num / 1000000000).toFixed(1).replace(/\.0$/, '') + 'G';
     }
     if (num >= 1000000) {
        return (num / 1000000).toFixed(1).replace(/\.0$/, '') + 'M';
     }
     if (num >= 1000) {
        return (num / 1000).toFixed(1).replace(/\.0$/, '') + 'K';
     }
     return num;
}

now nFormatter(33000) = 33K

现在nFormatter(33000)= 33 k

#4

Here's a simple solution that avoids all the if statements (with the power of Math).

这里有一个简单的解决方案，它避免了所有if语句(用数学的力量)。

var SI_PREFIXES = ["", "k", "M", "G", "T", "P", "E"];

function abbreviateNumber(number){

    // what tier? (determines SI prefix)
    var tier = Math.log10(number) / 3 | 0;

    // if zero, we don't need a prefix
    if(tier == 0) return number;

    // get prefix and determine scale
    var prefix = SI_PREFIXES[tier];
    var scale = Math.pow(10, tier * 3);

    // scale the number
    var scaled = number / scale;

    // format number and add prefix as suffix
    return scaled.toFixed(1) + prefix;
}

#5

/**
 * Shorten number to thousands, millions, billions, etc.
 * http://en.wikipedia.org/wiki/Metric_prefix
 *
 * @param {number} num Number to shorten.
 * @param {number} [digits=0] The number of digits to appear after the decimal point.
 * @returns {string|number}
 *
 * @example
 * // returns '12.5k'
 * shortenLargeNumber(12543, 1)
 *
 * @example
 * // returns '-13k'
 * shortenLargeNumber(-12567)
 *
 * @example
 * // returns '51M'
 * shortenLargeNumber(51000000)
 *
 * @example
 * // returns 651
 * shortenLargeNumber(651)
 *
 * @example
 * // returns 0.12345
 * shortenLargeNumber(0.12345)
 */
function shortenLargeNumber(num, digits) {
    var units = ['k', 'M', 'G', 'T', 'P', 'E', 'Z', 'Y'],
        decimal;

    for(var i=units.length-1; i>=0; i--) {
        decimal = Math.pow(1000, i+1);

        if(num <= -decimal || num >= decimal) {
            return +(num / decimal).toFixed(digits) + units[i];
        }
    }

    return num;
}

Thx @Cos for comment, I removed Math.round10 dependency.

@ Thx @Cos，我删除了Math。round10依赖。

#6

Give Credit to Waylon Flinn if you like this

This was improved from his more elegant approach to handle negative numbers and ".0" case.

这是从他处理负数和“。0”的情况。

The fewer loops and "if" cases you have, the better IMO.

循环次数越少，“如果”的情况就越好。

function abbreviateNumber(number) {
    var SI_POSTFIXES = ["", "k", "M", "G", "T", "P", "E"];
    var tier = Math.log10(Math.abs(number)) / 3 | 0;
    if(tier == 0) return number;
    var postfix = SI_POSTFIXES[tier];
    var scale = Math.pow(10, tier * 3);
    var scaled = number / scale;
    var formatted = scaled.toFixed(1) + '';
    if (/\.0$/.test(formatted))
      formatted = formatted.substr(0, formatted.length - 2);
    return formatted + postfix;
}

jsFiddle with test cases -> https://jsfiddle.net/xyug4nvz/7/

jsFiddle with测试用例—> https://jsfiddle.net/xyug4nvz/7/

#7

this is is quite elegant.

这很优雅。

function formatToUnits(number, precision) {
  const abbrev = ['', 'k', 'm', 'b', 't'];
  const unrangifiedOrder = Math.floor(Math.log10(Math.abs(number)) / 3)
  const order = Math.max(0, Math.min(unrangifiedOrder, abbrev.length -1 ))
  const suffix = abbrev[order];

  return (number / Math.pow(10, order * 3)).toFixed(precision) + suffix;
}

formatToUnits(12345, 2)
==> "12.35k"
formatToUnits(0, 3)
==> "0.000"

#8

Further improving @Yash's answer with negative number support:

进一步改善@Yash的答案，并提供负数支持:

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"

#9

You can use the d3-format package modeled after Python Advanced String Formatting PEP3101 :

您可以使用模仿Python高级字符串格式PEP3101的d3格式包:

var f = require('d3-format')
console.log(f.format('.2s')(2500)) // displays "2.5k"

#10

This post is quite old but I somehow reached to this post searching for something. SO to add my input Numeral js is the one stop solution now a days. It gives a large number of methods to help formatting the numbers

这篇文章很老了，但我还是找到了这篇文章。所以要添加输入数字js是一种停止方案。它提供了大量的方法来帮助格式化数字。

http://numeraljs.com/

http://numeraljs.com/

#11

Adding on the top answer, this will give 1k for 1000 instead of 1.0k

加上上面的答案，这将给出1000而不是1。0k的1k。

function kFormatter(num) {
    return num > 999 ? num % 1000 === 0 ? (num/1000).toFixed(0) + 'k' : (num/1000).toFixed(1) + 'k' : num
}

#12

/*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;
    }