如何在 JavaScript 中四舍五入？

Question

I am trying to round large digits. For instance, if I have this number:

12,645,982

I want to round this number and display it as:

13 mil

Or, if I have this number:

1,345

I want to round it and display it as:

1 thousand

How do I do this in JavaScript or jQuery?

Answer 1

Here is a utility function to format thousands, millions, and billions:

function MoneyFormat(labelValue) 
  {
  // Nine Zeroes for Billions
  return Math.abs(Number(labelValue)) >= 1.0e+9

       ? Math.abs(Number(labelValue)) / 1.0e+9 + "B"
       // Six Zeroes for Millions 
       : Math.abs(Number(labelValue)) >= 1.0e+6

       ? Math.abs(Number(labelValue)) / 1.0e+6 + "M"
       // Three Zeroes for Thousands
       : Math.abs(Number(labelValue)) >= 1.0e+3

       ? Math.abs(Number(labelValue)) / 1.0e+3 + "K"

       : Math.abs(Number(labelValue));

   }

Usage:

   var foo = MoneyFormat(1355);
   //Reformat result to one decimal place
   console.log(parseFloat(foo).toPrecision(2) + foo.replace(/[^B|M|K]/g,""))

References

• ECMAScript-5: Annex A

• Displaying numbers in JavaScript

Answer 2

Numeral JS .. If someone checks this out please check numeral Js. you just have to include the script and then its just one lineof code

numeral(yourNumber).format('0.0a')

Answer 3

var lazyround = function (num) {
    var parts = num.split(",");
    return parts.length > 1 ? (Math.round(parseInt(parts.join(""), 10) / Math.pow(1000, parts.length-1)) + " " + ["thousand", "million", "billion"][parts.length-2]) : parts[0];
};

alert(lazyround("9,012,345,678"));
alert(lazyround("12,345,678"));
alert(lazyround("345,678"));
alert(lazyround("678"));

it outputs this:

9 billion
12 million
346 thousand
678

have fun. this works fine and since i dont see that you did anything yourself this is obfuscated.

there you have a working example in jsfiddle: jsfiddle.net/p8pfB/

Answer 4

@Paul's answer is great, but it gives 773.282600918 M etc.

We probably dont want it for big numbers. Fixed using Math.round()

function moolah_round(num,locale='en') {
    // Nine Zeroes for Billions
    return Math.abs(Number(num)) >= 1.0e+9
        ? Math.round(Math.abs(Number(num)) / 1.0e+9 ) + " B"
        // Six Zeroes for Millions
        : Math.abs(Number(num)) >= 1.0e+6
            ? Math.round(Math.abs(Number(num)) / 1.0e+6 ) + " M"
            // Three Zeroes for Thousands
            : Math.abs(Number(num)) >= 1.0e+3
                ? Math.round(Math.abs(Number(num)) / 1.0e+3 ) + " K"
                : Math.abs(Number(num)); 
}

Answer 5

var number = 1345;
var zeroCount = 3;
var roundedNumber = Math.round( number / Math.pow(10,zeroCount) )

Just adjust the zeroCount to the 3 for thousands, 6 for millions, etc. I'm assuming you can use if statements and just need some help with the math functions.

Answer 6

Paul Sweatte's answer is good

Deepak Thomas's answer is nice

My answer is way more flexible/customizable and easier to read, so here goes:

Code

 function NumbFormat(options) { let number = Math.abs(Number(options.number)); // Nine zeros for Billions if (Number(number) >= 1.0e+9) { return (number / 1.0e+9).toFixed(options.billion.decimal) + ` ${options.billion.unit}`; } // Six zeros for Millions if (Number(number) >= 1.0e+6) { return (number / 1.0e+6).toFixed(options.million.decimal) + ` ${options.million.unit}`; } // Thrhee zeros for Thousands if (Number(number) >= 1.0e+3) { return (number / 1.0e+3).toFixed(options.thousand.decimal) + ` ${options.thousand.unit}`; } return number; } console.log(NumbFormat({ 'number': 100000000, 'billion': { 'decimal': 1, 'unit': 'B', }, 'million': { 'decimal': 1, 'unit': 'M', }, 'thousand': { 'decimal': 1, 'unit': 'K', }, })); console.log(NumbFormat({ 'number': 100000000, 'billion': { 'decimal': 1, 'unit': 'B', }, 'million': { 'decimal': 1, 'unit': 'Million', }, 'thousand': { 'decimal': 1, 'unit': 'K', }, })); console.log(NumbFormat({ 'number': 100000000, 'billion': { 'decimal': 1, 'unit': 'B', }, 'million': { 'decimal': 0, 'unit': 'Million', }, 'thousand': { 'decimal': 1, 'unit': 'K', }, })); console.log(NumbFormat({ 'number': 100000000, 'billion': { 'decimal': 1, 'unit': 'B', }, 'million': { 'decimal': 0, 'unit': 'M', }, 'thousand': { 'decimal': 1, 'unit': 'K', }, }));

With the above code you are in control of how many decimals you want for billions, millions or thousands, you are also in control of which unit you wish to display :)

Answer 7

I modified cetinajero's and Deepak Thomas's awesome answers to also handle negative numbers. (and not use ternaries)

 const numRound = (number) => { let num = number; num = Number(num); const billions = num / 1.0e9; const millions = num / 1.0e6; const thousands = num / 1.0e3; if (Math.abs(num) >= 1.0e9 && Math.abs(billions) >= 100) { return `${Math.round(billions)}B` } if (Math.abs(num) >= 1.0e9 && Math.abs(billions) >= 10) { return `${billions.toFixed(1)}B` } if (Math.abs(num) >= 1.0e9) { return `${billions.toFixed(2)}B` } if (Math.abs(num) >= 1.0e6 && Math.abs(millions) >= 100) { return `${Math.round(millions)}M` } if (Math.abs(num) >= 1.0e6 && Math.abs(millions) >= 10) { return `${millions.toFixed(1)}M` } if (Math.abs(num) >= 1.0e6) { return `${millions.toFixed(2)}M` } if (Math.abs(num) >= 1.0e3 && Math.abs(thousands) >= 100) { return `${Math.round(thousands)}K` } if (num >= 1.0e3 && thousands >= 10) { return `${thousands.toFixed(1)}K` } if (Math.abs(num) >= 1.0e3) { return `${thousands.toFixed(1)}K` } return num.toFixed(); }; console.log(numRound(-23400006)) console.log(numRound(1200)) console.log(numRound(654321)) console.log(numRound(12334000060))

Answer 8

Use str.length and the switch case

var number=12345;

something like this

switch (number.length) {
  case 4:
    alert(number/10000+"Thousand");
    break;

}

Answer 9

This is an improvement over Deepak Thomas's answer.

It handles decimals whenever it's necessary and gives more readability and control over the function.

Run the code snippet to see a 5 alert live demo showing its usage.

 function numRound(num) { num = Math.abs(Number(num)) const billions = num/1.0e+9 const millions = num/1.0e+6 const thousands = num/1.0e+3 return num >= 1.0e+9 && billions >= 100 ? Math.round(billions) + "B" : num >= 1.0e+9 && billions >= 10 ? billions.toFixed(1) + "B" : num >= 1.0e+9 ? billions.toFixed(2) + "B" : num >= 1.0e+6 && millions >= 100 ? Math.round(millions) + "M" : num >= 1.0e+6 && millions >= 10 ? millions.toFixed(1) + "M" : num >= 1.0e+6 ? millions.toFixed(2) + "M" : num >= 1.0e+3 && thousands >= 100 ? Math.round(thousands) + "K" : num >= 1.0e+3 && thousands >= 10 ? thousands.toFixed(1) + "K" : num >= 1.0e+3 ? thousands.toFixed(2) + "K" : num.toFixed() } alert(numRound(1234)) alert(numRound(23456)) alert(numRound(345678)) alert(numRound(4567890)) alert(numRound(56789012))

Answer 10

the simplest way to achieve this is to use Intl api.

example:

 const formatter = Intl.NumberFormat('en', {notation: 'compact'}) const K = formatter.format(1555) // thousand const M = formatter.format(1555555) // million const B = formatter.format(1555555555) // billion const T = formatter.format(1555555555555) // trillion console.log(K, M, B, T)