Suppose I have a value of 15.7784514, I want to display it 15.77 with no rounding.
var num = parseFloat(15.7784514);
document.write(num.toFixed(1)+"<br />");
document.write(num.toFixed(2)+"<br />");
document.write(num.toFixed(3)+"<br />");
document.write(num.toFixed(10));
Results in -
15.8
15.78
15.778
15.7784514000
How do I display 15.77?
Convert the number into a string, match the number up to the second decimal place:
function calc(theform) {
var num = theform.original.value, rounded = theform.rounded
var with2Decimals = num.toString().match(/^-?\d+(?:\.\d{0,2})?/)[0]
rounded.value = with2Decimals
}
<form onsubmit="return calc(this)">
Original number: <input name="original" type="text" onkeyup="calc(form)" onchange="calc(form)" />
<br />"Rounded" number: <input name="rounded" type="text" placeholder="readonly" readonly>
</form>
The toFixed method fails in some cases unlike toString, so be very careful with it.
Update 5 Nov 2016
New answer, always accurate
function toFixed(num, fixed) {
var re = new RegExp('^-?\\d+(?:\.\\d{0,' + (fixed || -1) + '})?');
return num.toString().match(re)[0];
}
As floating point math in javascript will always have edge cases, the previous solution will be accurate most of the time which is not good enough.
There are some solutions to this like num.toPrecision, BigDecimal.js, and accounting.js.
Yet, I believe that merely parsing the string will be the simplest and always accurate.
Basing the update on the well written regex from the accepted answer by #Gumbo, this new toFixed function will always work as expected.
Old answer, not always accurate.
Roll your own toFixed function:
function toFixed(num, fixed) {
fixed = fixed || 0;
fixed = Math.pow(10, fixed);
return Math.floor(num * fixed) / fixed;
}
Another single-line solution :
number = Math.trunc(number*100)/100
I used 100 because you want to truncate to the second digit, but a more flexible solution would be :
number = Math.trunc(number*Math.pow(10, digits))/Math.pow(10, digits)
where digits is the amount of decimal digits to keep.
See Math.trunc specs for details and browser compatibility.
I opted to write this instead to manually remove the remainder with strings so I don't have to deal with the math issues that come with numbers:
num = num.toString(); //If it's not already a String
num = num.slice(0, (num.indexOf("."))+3); //With 3 exposing the hundredths place
Number(num); //If you need it back as a Number
This will give you "15.77" with num = 15.7784514;
Update (Jan 2021)
Depending on its range, a number in javascript may be shown in scientific notation. For example, if you type 0.0000001 in the console, you may see it as 1e-7, whereas 0.000001 appears unchanged (0.000001).
If your application works on a range of numbers for which scientific notation is not involved, you can just ignore this update and use the original answer below.
This update is about adding a function that checks if the number is in scientific format and, if so, converts it into decimal format. Here I'm proposing this one, but you can use any other function that achieves the same goal, according to your application's needs:
function toFixed(x) {
if (Math.abs(x) < 1.0) {
let e = parseInt(x.toString().split('e-')[1]);
if (e) {
x *= Math.pow(10,e-1);
x = '0.' + (new Array(e)).join('0') + x.toString().substring(2);
}
} else {
let e = parseInt(x.toString().split('+')[1]);
if (e > 20) {
e -= 20;
x /= Math.pow(10,e);
x += (new Array(e+1)).join('0');
}
}
return x;
}
Now just apply that function to the parameter (that's the only change with respect to the original answer):
function toFixedTrunc(x, n) {
x = toFixed(x)
// From here on the code is the same than the original answer
const v = (typeof x === 'string' ? x : x.toString()).split('.');
if (n <= 0) return v[0];
let f = v[1] || '';
if (f.length > n) return `${v[0]}.${f.substr(0,n)}`;
while (f.length < n) f += '0';
return `${v[0]}.${f}`
}
This updated version addresses also a case mentioned in a comment:
toFixedTrunc(0.000000199, 2) => "0.00"
Again, choose what fits your application needs at best.
Original answer (October 2017)
General solution to truncate (no rounding) a number to the n-th decimal digit and convert it to a string with exactly n decimal digits, for any n≥0.
function toFixedTrunc(x, n) {
const v = (typeof x === 'string' ? x : x.toString()).split('.');
if (n <= 0) return v[0];
let f = v[1] || '';
if (f.length > n) return `${v[0]}.${f.substr(0,n)}`;
while (f.length < n) f += '0';
return `${v[0]}.${f}`
}
where x can be either a number (which gets converted into a string) or a string.
Here are some tests for n=2 (including the one requested by OP):
0 => 0.00
0.01 => 0.01
0.5839 => 0.58
0.999 => 0.99
1.01 => 1.01
2 => 2.00
2.551 => 2.55
2.99999 => 2.99
4.27 => 4.27
15.7784514 => 15.77
123.5999 => 123.59
And for some other values of n:
15.001097 => 15.0010 (n=4)
0.000003298 => 0.0000032 (n=7)
0.000003298257899 => 0.000003298257 (n=12)
parseInt is faster then Math.floor
function floorFigure(figure, decimals){
if (!decimals) decimals = 2;
var d = Math.pow(10,decimals);
return (parseInt(figure*d)/d).toFixed(decimals);
};
floorFigure(123.5999) => "123.59"
floorFigure(123.5999, 3) => "123.599"
num = 19.66752
f = num.toFixed(3).slice(0,-1)
alert(f)
This will return 19.66
Simple do this
number = parseInt(number * 100)/100;
Just truncate the digits:
function truncDigits(inputNumber, digits) {
const fact = 10 ** digits;
return Math.floor(inputNumber * fact) / fact;
}
This is not a safe alternative, as many others commented examples with numbers that turn into exponential notation, that scenery is not covered by this function
// typescript
// function formatLimitDecimals(value: number, decimals: number): number {
function formatLimitDecimals(value, decimals) {
const stringValue = value.toString();
if(stringValue.includes('e')) {
// TODO: remove exponential notation
throw 'invald number';
} else {
const [integerPart, decimalPart] = stringValue.split('.');
if(decimalPart) {
return +[integerPart, decimalPart.slice(0, decimals)].join('.')
} else {
return integerPart;
}
}
}
console.log(formatLimitDecimals(4.156, 2)); // 4.15
console.log(formatLimitDecimals(4.156, 8)); // 4.156
console.log(formatLimitDecimals(4.156, 0)); // 4
console.log(formatLimitDecimals(0, 4)); // 0
// not covered
console.log(formatLimitDecimals(0.000000199, 2)); // 0.00
These solutions do work, but to me seem unnecessarily complicated. I personally like to use the modulus operator to obtain the remainder of a division operation, and remove that. Assuming that num = 15.7784514:
num-=num%.01;
This is equivalent to saying num = num - (num % .01).
I fixed using following simple way-
var num = 15.7784514;
Math.floor(num*100)/100;
Results will be 15.77
My version for positive numbers:
function toFixed_norounding(n,p)
{
var result = n.toFixed(p);
return result <= n ? result: (result - Math.pow(0.1,p)).toFixed(p);
}
Fast, pretty, obvious. (version for positive numbers)
The answers here didn't help me, it kept rounding up or giving me the wrong decimal.
my solution converts your decimal to a string, extracts the characters and then returns the whole thing as a number.
function Dec2(num) {
num = String(num);
if(num.indexOf('.') !== -1) {
var numarr = num.split(".");
if (numarr.length == 1) {
return Number(num);
}
else {
return Number(numarr[0]+"."+numarr[1].charAt(0)+numarr[1].charAt(1));
}
}
else {
return Number(num);
}
}
Dec2(99); // 99
Dec2(99.9999999); // 99.99
Dec2(99.35154); // 99.35
Dec2(99.8); // 99.8
Dec2(10265.985475); // 10265.98
The following code works very good for me:
num.toString().match(/.\*\\..{0,2}|.\*/)[0];
This worked well for me. I hope it will fix your issues too.
function toFixedNumber(number) {
const spitedValues = String(number.toLocaleString()).split('.');
let decimalValue = spitedValues.length > 1 ? spitedValues[1] : '';
decimalValue = decimalValue.concat('00').substr(0,2);
return '$'+spitedValues[0] + '.' + decimalValue;
}
// 5.56789 ----> $5.56
// 0.342 ----> $0.34
// -10.3484534 ----> $-10.34
// 600 ----> $600.00
function convertNumber(){
var result = toFixedNumber(document.getElementById("valueText").value);
document.getElementById("resultText").value = result;
}
function toFixedNumber(number) {
const spitedValues = String(number.toLocaleString()).split('.');
let decimalValue = spitedValues.length > 1 ? spitedValues[1] : '';
decimalValue = decimalValue.concat('00').substr(0,2);
return '$'+spitedValues[0] + '.' + decimalValue;
}
<div>
<input type="text" id="valueText" placeholder="Input value here..">
<br>
<button onclick="convertNumber()" >Convert</button>
<br><hr>
<input type="text" id="resultText" placeholder="result" readonly="true">
</div>
An Easy way to do it is the next but is necessary ensure that the amount parameter is given as a string.
function truncate(amountAsString, decimals = 2){
var dotIndex = amountAsString.indexOf('.');
var toTruncate = dotIndex !== -1 && ( amountAsString.length > dotIndex + decimals + 1);
var approach = Math.pow(10, decimals);
var amountToTruncate = toTruncate ? amountAsString.slice(0, dotIndex + decimals +1) : amountAsString;
return toTruncate
? Math.floor(parseFloat(amountToTruncate) * approach ) / approach
: parseFloat(amountAsString);
}
console.log(truncate("7.99999")); //OUTPUT ==> 7.99
console.log(truncate("7.99999", 3)); //OUTPUT ==> 7.999
console.log(truncate("12.799999999999999")); //OUTPUT ==> 7.99
Here you are. An answer that shows yet another way to solve the problem:
// For the sake of simplicity, here is a complete function:
function truncate(numToBeTruncated, numOfDecimals) {
var theNumber = numToBeTruncated.toString();
var pointIndex = theNumber.indexOf('.');
return +(theNumber.slice(0, pointIndex > -1 ? ++numOfDecimals + pointIndex : undefined));
}
Note the use of + before the final expression. That is to convert our truncated, sliced string back to number type.
Hope it helps!
truncate without zeroes
function toTrunc(value,n){
return Math.floor(value*Math.pow(10,n))/(Math.pow(10,n));
}
or
function toTrunc(value,n){
x=(value.toString()+".0").split(".");
return parseFloat(x[0]+"."+x[1].substr(0,n));
}
test:
toTrunc(17.4532,2) //17.45
toTrunc(177.4532,1) //177.4
toTrunc(1.4532,1) //1.4
toTrunc(.4,2) //0.4
truncate with zeroes
function toTruncFixed(value,n){
return toTrunc(value,n).toFixed(n);
}
test:
toTrunc(17.4532,2) //17.45
toTrunc(177.4532,1) //177.4
toTrunc(1.4532,1) //1.4
toTrunc(.4,2) //0.40
If you exactly wanted to truncate to 2 digits of precision, you can go with a simple logic:
function myFunction(number) {
var roundedNumber = number.toFixed(2);
if (roundedNumber > number)
{
roundedNumber = roundedNumber - 0.01;
}
return roundedNumber;
}
I used (num-0.05).toFixed(1) to get the second decimal floored.
It's more reliable to get two floating points without rounding.
Reference Answer
var number = 10.5859;
var fixed2FloatPoints = parseInt(number * 100) / 100;
console.log(fixed2FloatPoints);
Thank You !
My solution in typescript (can easily be ported to JS):
/**
* Returns the price with correct precision as a string
*
* #param price The price in decimal to be formatted.
* #param decimalPlaces The number of decimal places to use
* #return string The price in Decimal formatting.
*/
type toDecimal = (price: number, decimalPlaces?: number) => string;
const toDecimalOdds: toDecimal = (
price: number,
decimalPlaces: number = 2,
): string => {
const priceString: string = price.toString();
const pointIndex: number = priceString.indexOf('.');
// Return the integer part if decimalPlaces is 0
if (decimalPlaces === 0) {
return priceString.substr(0, pointIndex);
}
// Return value with 0s appended after decimal if the price is an integer
if (pointIndex === -1) {
const padZeroString: string = '0'.repeat(decimalPlaces);
return `${priceString}.${padZeroString}`;
}
// If numbers after decimal are less than decimalPlaces, append with 0s
const padZeroLen: number = priceString.length - pointIndex - 1;
if (padZeroLen > 0 && padZeroLen < decimalPlaces) {
const padZeroString: string = '0'.repeat(padZeroLen);
return `${priceString}${padZeroString}`;
}
return priceString.substr(0, pointIndex + decimalPlaces + 1);
};
Test cases:
expect(filters.toDecimalOdds(3.14159)).toBe('3.14');
expect(filters.toDecimalOdds(3.14159, 2)).toBe('3.14');
expect(filters.toDecimalOdds(3.14159, 0)).toBe('3');
expect(filters.toDecimalOdds(3.14159, 10)).toBe('3.1415900000');
expect(filters.toDecimalOdds(8.2)).toBe('8.20');
Any improvements?
Another solution, that truncates and round:
function round (number, decimals, truncate) {
if (truncate) {
number = number.toFixed(decimals + 1);
return parseFloat(number.slice(0, -1));
}
var n = Math.pow(10.0, decimals);
return Math.round(number * n) / n;
};
function limitDecimalsWithoutRounding(val, decimals){
let parts = val.toString().split(".");
return parseFloat(parts[0] + "." + parts[1].substring(0, decimals));
}
var num = parseFloat(15.7784514);
var new_num = limitDecimalsWithoutRounding(num, 2);
Roll your own toFixed function: for positive values Math.floor works fine.
function toFixed(num, fixed) {
fixed = fixed || 0;
fixed = Math.pow(10, fixed);
return Math.floor(num * fixed) / fixed;
}
For negative values Math.floor is round of the values. So you can use Math.ceil instead.
Example,
Math.ceil(-15.778665 * 10000) / 10000 = -15.7786
Math.floor(-15.778665 * 10000) / 10000 = -15.7787 // wrong.
Gumbo's second solution, with the regular expression, does work but is slow because of the regular expression. Gumbo's first solution fails in certain situations due to imprecision in floating points numbers. See the JSFiddle for a demonstration and a benchmark. The second solution takes about 1636 nanoseconds per call on my current system, Intel Core i5-2500 CPU at 3.30 GHz.
The solution I've written involves adding a small compensation to take care of floating point imprecision. It is basically instantaneous, i.e. on the order of nanoseconds. I clocked 2 nanoseconds per call but the JavaScript timers are not very precise or granular. Here is the JS Fiddle and the code.
function toFixedWithoutRounding (value, precision)
{
var factorError = Math.pow(10, 14);
var factorTruncate = Math.pow(10, 14 - precision);
var factorDecimal = Math.pow(10, precision);
return Math.floor(Math.floor(value * factorError + 1) / factorTruncate) / factorDecimal;
}
var values = [1.1299999999, 1.13, 1.139999999, 1.14, 1.14000000001, 1.13 * 100];
for (var i = 0; i < values.length; i++)
{
var value = values[i];
console.log(value + " --> " + toFixedWithoutRounding(value, 2));
}
for (var i = 0; i < values.length; i++)
{
var value = values[i];
console.log(value + " --> " + toFixedWithoutRounding(value, 4));
}
console.log("type of result is " + typeof toFixedWithoutRounding(1.13 * 100 / 100, 2));
// Benchmark
var value = 1.13 * 100;
var startTime = new Date();
var numRun = 1000000;
var nanosecondsPerMilliseconds = 1000000;
for (var run = 0; run < numRun; run++)
toFixedWithoutRounding(value, 2);
var endTime = new Date();
var timeDiffNs = nanosecondsPerMilliseconds * (endTime - startTime);
var timePerCallNs = timeDiffNs / numRun;
console.log("Time per call (nanoseconds): " + timePerCallNs);
Building on David D's answer:
function NumberFormat(num,n) {
var num = (arguments[0] != null) ? arguments[0] : 0;
var n = (arguments[1] != null) ? arguments[1] : 2;
if(num > 0){
num = String(num);
if(num.indexOf('.') !== -1) {
var numarr = num.split(".");
if (numarr.length > 1) {
if(n > 0){
var temp = numarr[0] + ".";
for(var i = 0; i < n; i++){
if(i < numarr[1].length){
temp += numarr[1].charAt(i);
}
}
num = Number(temp);
}
}
}
}
return Number(num);
}
console.log('NumberFormat(123.85,2)',NumberFormat(123.85,2));
console.log('NumberFormat(123.851,2)',NumberFormat(123.851,2));
console.log('NumberFormat(0.85,2)',NumberFormat(0.85,2));
console.log('NumberFormat(0.851,2)',NumberFormat(0.851,2));
console.log('NumberFormat(0.85156,2)',NumberFormat(0.85156,2));
console.log('NumberFormat(0.85156,4)',NumberFormat(0.85156,4));
console.log('NumberFormat(0.85156,8)',NumberFormat(0.85156,8));
console.log('NumberFormat(".85156",2)',NumberFormat(".85156",2));
console.log('NumberFormat("0.85156",2)',NumberFormat("0.85156",2));
console.log('NumberFormat("1005.85156",2)',NumberFormat("1005.85156",2));
console.log('NumberFormat("0",2)',NumberFormat("0",2));
console.log('NumberFormat("",2)',NumberFormat("",2));
console.log('NumberFormat(85156,8)',NumberFormat(85156,8));
console.log('NumberFormat("85156",2)',NumberFormat("85156",2));
console.log('NumberFormat("85156.",2)',NumberFormat("85156.",2));
// NumberFormat(123.85,2) 123.85
// NumberFormat(123.851,2) 123.85
// NumberFormat(0.85,2) 0.85
// NumberFormat(0.851,2) 0.85
// NumberFormat(0.85156,2) 0.85
// NumberFormat(0.85156,4) 0.8515
// NumberFormat(0.85156,8) 0.85156
// NumberFormat(".85156",2) 0.85
// NumberFormat("0.85156",2) 0.85
// NumberFormat("1005.85156",2) 1005.85
// NumberFormat("0",2) 0
// NumberFormat("",2) 0
// NumberFormat(85156,8) 85156
// NumberFormat("85156",2) 85156
// NumberFormat("85156.",2) 85156
Already there are some suitable answer with regular expression and arithmetic calculation, you can also try this
function myFunction() {
var str = 12.234556;
str = str.toString().split('.');
var res = str[1].slice(0, 2);
document.getElementById("demo").innerHTML = str[0]+'.'+res;
}
// output: 12.23
Here is what is did it with string
export function withoutRange(number) {
const str = String(number);
const dotPosition = str.indexOf('.');
if (dotPosition > 0) {
const length = str.substring().length;
const end = length > 3 ? 3 : length;
return str.substring(0, dotPosition + end);
}
return str;
}
I'm generating a random number with the code below:
Math.floor((Math.random() * 9999) * 7);
Some of the results I'm getting:
45130,
2611,
34509,
36658
How would I get results like this(with 2 letters included):
TT45130,
PO2611,
KL34509,
GH36658
Side question:
What is the range of numbers that Math.random() carries? Can I set a specific range of values? Not necessary to answer but just curious.
You can use a function like below to get a random uppercase character:
function getRandomUppercaseChar() {
var r = Math.floor(Math.random() * 26);
return String.fromCharCode(65 + r);
}
So to generate a code as you specified with a two-letter prefix:
function generateCode() {
var prefix = new Array(2).fill().map(() => getRandomUppercaseChar()).join(""),
integer = Math.floor((Math.random() * 9999) * 7);
return prefix + integer;
}
NOTE: The above generateCode function uses modern ES6 and ES5 javascript, which is perfectly fine in a modern environment (such as Node.js or a current browser). However, if you wanted greater compatibility (for example, to ensure that it works in old browsers), you could rewrite it like so:
function generateCode() {
var integer = Math.floor((Math.random() * 9999) * 7);
for (var i = 0, prefix = ""; i < 2; ++i)
prefix += getRandomUppercaseChar();
return prefix + integer;
}
Try the simpler answer
var randomNumber = function () {
return Math.floor((Math.random() * 9999) * 7);
}
var randomChar = function () {
return String.fromCharCode(64 + Math.floor((Math.random() * 26)+1));
}
console.log(randomChar()+randomChar()+randomNumber());
//Sample outputs
HB10527 DR25496 IJ12394
Or you can use Number#toString for this purpose with radix = 36.
function getRChar() {
return (Math.random() * 26 + 10 | 0).toString(36).toUpperCase();
}
var s = getRChar() + getRChar() + Math.floor((Math.random() * 9999) * 7);
document.write(s);
If you need to generate a random string with JS, the most common way is to define an alphabet and pick random indices from that:
var alphabet = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
var numbers = "0123456789";
var randomString = "";
// Pick two random chars
for (var i = 0; i < 2; i++) {
var rand = Math.floor(Math.random()*alphabet.length);
randomString = randomString + alphabet.charAt(rand);
}
// Pick four random digits
for (var i = 0; i < 4; i++) {
var rand = Math.floor(Math.random()*numbers.length);
randomString = randomString + numbers.charAt(rand);
}
// randomString now contains the string you want
Sample strings:
OJ8225
YL5053
BD7911
ES0159
You could use String.fromCharCode() with a random integer between 65 and 90 to get an uppercase letter, i.e.
String.fromCharCode(Math.random() * 26 + 65) + String.fromCharCode(Math.random() * 26 + 65) + Math.floor((Math.random() * 9999) * 7);
gives med the results: "SH21248", "BY42401", "TD35918".
If you want to guarantee that the string always has the same length, you could also use
String.fromCharCode(Math.random() * 26 + 65) + String.fromCharCode(Math.random() * 26 + 65) + Math.floor(Math.random() * 59993 + 10000);
Math.random() always returns a number between 0 and 1, but never 0 or 1 exactly.
An array of the alphabet, a random number is generated to get a random letter, repeated to get a second random letter and then joined to the random number generated as in your code:
var alphabet=['A','B','C','D','E','F','G','H','I','J','K','L','M','N','O','P','Q','R','S','T','U','V','W','X','Y','Z'];
var ranletter1 = alphabet[Math.floor(Math.random() * alphabet.length)];
var ranletter2 = alphabet[Math.floor(Math.random() * alphabet.length)];
var ranNum = Math.floor((Math.random() * 9999) * 7);
var ranCode = ranletter1 + ranletter2+ ranNum;
In JavaScript, when converting from a float to a string, how can I get just 2 digits after the decimal point? For example, 0.34 instead of 0.3445434.
There are functions to round numbers. For example:
var x = 5.0364342423;
print(x.toFixed(2));
will print 5.04.
EDIT:
Fiddle
var result = Math.round(original*100)/100;
The specifics, in case the code isn't self-explanatory.
edit: ...or just use toFixed, as proposed by Tim Büthe. Forgot that one, thanks (and an upvote) for reminder :)
Be careful when using toFixed():
First, rounding the number is done using the binary representation of the number, which might lead to unexpected behaviour. For example
(0.595).toFixed(2) === '0.59'
instead of '0.6'.
Second, there's an IE bug with toFixed(). In IE (at least up to version 7, didn't check IE8), the following holds true:
(0.9).toFixed(0) === '0'
It might be a good idea to follow kkyy's suggestion or to use a custom toFixed() function, eg
function toFixed(value, precision) {
var power = Math.pow(10, precision || 0);
return String(Math.round(value * power) / power);
}
One more problem to be aware of, is that toFixed() can produce unnecessary zeros at the end of the number.
For example:
var x=(23-7.37)
x
15.629999999999999
x.toFixed(6)
"15.630000"
The idea is to clean up the output using a RegExp:
function humanize(x){
return x.toFixed(6).replace(/\.?0*$/,'');
}
The RegExp matches the trailing zeros (and optionally the decimal point) to make sure it looks good for integers as well.
humanize(23-7.37)
"15.63"
humanize(1200)
"1200"
humanize(1200.03)
"1200.03"
humanize(3/4)
"0.75"
humanize(4/3)
"1.333333"
var x = 0.3445434
x = Math.round (x*100) / 100 // this will make nice rounding
The key here I guess is to round up correctly first, then you can convert it to String.
function roundOf(n, p) {
const n1 = n * Math.pow(10, p + 1);
const n2 = Math.floor(n1 / 10);
if (n1 >= (n2 * 10 + 5)) {
return (n2 + 1) / Math.pow(10, p);
}
return n2 / Math.pow(10, p);
}
// All edge cases listed in this thread
roundOf(95.345, 2); // 95.35
roundOf(95.344, 2); // 95.34
roundOf(5.0364342423, 2); // 5.04
roundOf(0.595, 2); // 0.60
roundOf(0.335, 2); // 0.34
roundOf(0.345, 2); // 0.35
roundOf(551.175, 2); // 551.18
roundOf(0.3445434, 2); // 0.34
Now you can safely format this value with toFixed(p).
So with your specific case:
roundOf(0.3445434, 2).toFixed(2); // 0.34
There is a problem with all those solutions floating around using multipliers. Both kkyy and Christoph's solutions are wrong unfortunately.
Please test your code for number 551.175 with 2 decimal places - it will round to 551.17 while it should be 551.18 ! But if you test for ex. 451.175 it will be ok - 451.18. So it's difficult to spot this error at a first glance.
The problem is with multiplying: try 551.175 * 100 = 55117.49999999999 (ups!)
So my idea is to treat it with toFixed() before using Math.round();
function roundFix(number, precision)
{
var multi = Math.pow(10, precision);
return Math.round( (number * multi).toFixed(precision + 1) ) / multi;
}
If you want the string without round you can use this RegEx (maybe is not the most efficient way... but is really easy)
(2.34567778).toString().match(/\d+\.\d{2}/)[0]
// '2.34'
function trimNumber(num, len) {
const modulu_one = 1;
const start_numbers_float=2;
var int_part = Math.trunc(num);
var float_part = String(num % modulu_one);
float_part = float_part.slice(start_numbers_float, start_numbers_float+len);
return int_part+'.'+float_part;
}
There is no way to avoid inconsistent rounding for prices with x.xx5 as actual value using either multiplication or division. If you need to calculate correct prices client-side you should keep all amounts in cents. This is due to the nature of the internal representation of numeric values in JavaScript. Notice that Excel suffers from the same problems so most people wouldn't notice the small errors caused by this phenomen. However errors may accumulate whenever you add up a lot of calculated values, there is a whole theory around this involving the order of calculations and other methods to minimize the error in the final result. To emphasize on the problems with decimal values, please note that 0.1 + 0.2 is not exactly equal to 0.3 in JavaScript, while 1 + 2 is equal to 3.
Maybe you'll also want decimal separator? Here is a function I just made:
function formatFloat(num,casasDec,sepDecimal,sepMilhar) {
if (num < 0)
{
num = -num;
sinal = -1;
} else
sinal = 1;
var resposta = "";
var part = "";
if (num != Math.floor(num)) // decimal values present
{
part = Math.round((num-Math.floor(num))*Math.pow(10,casasDec)).toString(); // transforms decimal part into integer (rounded)
while (part.length < casasDec)
part = '0'+part;
if (casasDec > 0)
{
resposta = sepDecimal+part;
num = Math.floor(num);
} else
num = Math.round(num);
} // end of decimal part
while (num > 0) // integer part
{
part = (num - Math.floor(num/1000)*1000).toString(); // part = three less significant digits
num = Math.floor(num/1000);
if (num > 0)
while (part.length < 3) // 123.023.123 if sepMilhar = '.'
part = '0'+part; // 023
resposta = part+resposta;
if (num > 0)
resposta = sepMilhar+resposta;
}
if (sinal < 0)
resposta = '-'+resposta;
return resposta;
}
/** don't spend 5 minutes, use my code **/
function prettyFloat(x,nbDec) {
if (!nbDec) nbDec = 100;
var a = Math.abs(x);
var e = Math.floor(a);
var d = Math.round((a-e)*nbDec); if (d == nbDec) { d=0; e++; }
var signStr = (x<0) ? "-" : " ";
var decStr = d.toString(); var tmp = 10; while(tmp<nbDec && d*tmp < nbDec) {decStr = "0"+decStr; tmp*=10;}
var eStr = e.toString();
return signStr+eStr+"."+decStr;
}
prettyFloat(0); // "0.00"
prettyFloat(-1); // "-1.00"
prettyFloat(-0.999); // "-1.00"
prettyFloat(0.5); // "0.50"
I use this code to format floats. It is based on toPrecision() but it strips unnecessary zeros. I would welcome suggestions for how to simplify the regex.
function round(x, n) {
var exp = Math.pow(10, n);
return Math.floor(x*exp + 0.5)/exp;
}
Usage example:
function test(x, n, d) {
var rounded = rnd(x, d);
var result = rounded.toPrecision(n);
result = result.replace(/\.?0*$/, '');
result = result.replace(/\.?0*e/, 'e');
result = result.replace('e+', 'e');
return result;
}
document.write(test(1.2000e45, 3, 2) + '=' + '1.2e45' + '<br>');
document.write(test(1.2000e+45, 3, 2) + '=' + '1.2e45' + '<br>');
document.write(test(1.2340e45, 3, 2) + '=' + '1.23e45' + '<br>');
document.write(test(1.2350e45, 3, 2) + '=' + '1.24e45' + '<br>');
document.write(test(1.0000, 3, 2) + '=' + '1' + '<br>');
document.write(test(1.0100, 3, 2) + '=' + '1.01' + '<br>');
document.write(test(1.2340, 4, 2) + '=' + '1.23' + '<br>');
document.write(test(1.2350, 4, 2) + '=' + '1.24' + '<br>');
countDecimals = value => {
if (Math.floor(value) === value) return 0;
let stringValue = value.toString().split(".")[1];
if (stringValue) {
return value.toString().split(".")[1].length
? value.toString().split(".")[1].length
: 0;
} else {
return 0;
}
};
formatNumber=(ans)=>{
let decimalPlaces = this.countDecimals(ans);
ans = 1 * ans;
if (decimalPlaces !== 0) {
let onePlusAns = ans + 1;
let decimalOnePlus = this.countDecimals(onePlusAns);
if (decimalOnePlus < decimalPlaces) {
ans = ans.toFixed(decimalPlaces - 1).replace(/\.?0*$/, "");
} else {
let tenMulAns = ans * 10;
let decimalTenMul = this.countDecimals(tenMulAns);
if (decimalTenMul + 1 < decimalPlaces) {
ans = ans.toFixed(decimalPlaces - 1).replace(/\.?0*$/, "");
}
}
}
}
I just add 1 to the value and count the decimal digits present in the original value and the added value. If I find the decimal digits after adding one less than the original decimal digits, I just call the toFixed() with (original decimals - 1). I also check by multiplying the original value by 10 and follow the same logic in case adding one doesn't reduce redundant decimal places.
A simple workaround to handle floating-point number rounding in JS. Works in most cases I tried.