# Truncate number to two decimal places without rounding

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;
}
``````

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;

# 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
0.000000199 => 1.99 *
``````

* As mentioned in the note, that's due to javascript implicit conversion into exponential for "1.99e-7" 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"
``````

Simple do this

``````number = parseInt(number * 100)/100;
``````
``````num = 19.66752
f = num.toFixed(3).slice(0,-1)
``````

This will return 19.66

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).

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
``````

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 following code works very good for me:

``````num.toString().match(/.\*\\..{0,2}|.\*/)[0];
``````

I fixed using following simple way-

``````var num = 15.7784514;
Math.floor(num*100)/100;
``````

Results will be 15.77

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){
}
``````

test:

``````toTrunc(17.4532,2)  //17.45
toTrunc(177.4532,1) //177.4
toTrunc(1.4532,1)   //1.4
toTrunc(.4,2)       //0.40
``````

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>``````

Just truncate the digits:

``````function truncDigits(inputNumber, digits) {
const fact = 10 ** digits;
return Math.floor(inputNumber * fact) / fact;
}
``````

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;
?  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
``````

``````function formatLimitDecimals(value, decimals) {
value = value.toString().split('.')

if (value.length === 2) {
return Number([value[0], value[1].slice(0, decimals)].join('.'))
} else {
return Number(value[0]);
}
}

console.log(formatLimitDecimals(4.156, 2)); // 4.15
console.log(formatLimitDecimals(4.156, 8)); // 4.156
console.log(formatLimitDecimals(4.156, 0)); // 4``````

I used `(num-0.05).toFixed(1)` to get the second decimal floored.

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.

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);

}

// If numbers after decimal are less than decimalPlaces, append with 0s
const padZeroLen: number = priceString.length - pointIndex - 1;
if (padZeroLen > 0 && padZeroLen < decimalPlaces) {

}

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?

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
``````

It's more reliable to get two floating points without rounding.

``````var number = 10.5859;
var fixed2FloatPoints = parseInt(number * 100) / 100;
console.log(fixed2FloatPoints);``````

Thank You !

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;
}
``````

All these answers seem a bit complicated. I would just subtract 0.5 from your number and use toFixed().

The most efficient solution (for 2 fraction digits) is to subtract 0.005 before calling `toFixed()`

``````function toFixed2( num ) { return (num-0.005).toFixed(2) }
``````

Negative numbers will be rounded down too (away from zero). The OP didn't tell anything about negative numbers.

I know there are already few working examples, but I think it's worth to propose my String.toFixed equivalent, some may find it helpful.

That's my toFixed alternative, that don't round numbers, just truncates it or adds zeroes according to given precision. For extra long number it uses JS build-in rounding when precision is to long. Function works for all problematic number I've found on stack.

``````function toFixedFixed(value, precision = 0) {
let stringValue = isNaN(+value) ? '0' : String(+value);

if (stringValue.indexOf('e') > -1 || stringValue === 'Infinity' || stringValue === '-Infinity') {
throw new Error('To large number to be processed');
}

let [ beforePoint, afterPoint ] = stringValue.indexOf('.') > -1 ? stringValue.split('.') : [ stringValue, ''];

// Force automatic rounding for some long real numbers that ends with 99X, by converting it to string, cutting off last digit, then adding extra nines and casting it on number again
// e.g. 2.0199999999999996: +('2.019999999999999' + '9999') will give 2.02
if (stringValue.length >= 17 && afterPoint.length > 2 && +afterPoint.substr(afterPoint.length - 3) > 995) {
stringValue = String(+(stringValue.substr(0, afterPoint.length - 1) + '9'.repeat(stringValue.split('.').shift().length + 4)));
[ beforePoint, afterPoint ] = String(stringValue).indexOf('.') > -1 ? stringValue.split('.') : [ stringValue, ''];
}

if (precision === 0) {
return beforePoint;
} else if (afterPoint.length > precision) {
return `\${beforePoint}.\${afterPoint.substr(0, precision)}`;
} else {
return `\${beforePoint}.\${afterPoint}\${'0'.repeat(precision - afterPoint.length)}`;
}
}
``````

Keep in mind that precision may not be handled properly for numbers of length 18 and greater, e.g. 64-bit Chrome will round it or add "e+" / "e-" to keep number's length at 18.

If you want to perform operations on real numbers it's safer to multiply it by `Math.sqrt(10, precision)` first, make the calculations, then dive result by multiplier's value.

Example:

`0.06 + 0.01` is `0.06999999999999999`, and every formatting function that's not rounding will truncate it to `0.06` for precision `2`.

But if you'll perform same operation with multiplier&divider: `(0.06 * 100 + 0.01 * 100) / 100`, you'll get `0.07`.

That's why it's so important to use such multiplier/divider when dealing with real numbers in javascript, especially when you're calculating money...