[javascript] 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?

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

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(){_x000D_
  var result = toFixedNumber(document.getElementById("valueText").value);_x000D_
  document.getElementById("resultText").value = result;_x000D_
}_x000D_
_x000D_
_x000D_
function toFixedNumber(number) {_x000D_
        const spitedValues = String(number.toLocaleString()).split('.');_x000D_
        let decimalValue = spitedValues.length > 1 ? spitedValues[1] : '';_x000D_
        decimalValue = decimalValue.concat('00').substr(0,2);_x000D_
_x000D_
        return '$'+spitedValues[0] + '.' + decimalValue;_x000D_
}

<div>_x000D_
  <input type="text" id="valueText" placeholder="Input value here..">_x000D_
  <br>_x000D_
  <button onclick="convertNumber()" >Convert</button>_x000D_
  <br><hr>_x000D_
  <input type="text" id="resultText" placeholder="result" readonly="true">_x000D_
</div>

const toFixed = (value) => value.toString().slice(0,5);

Thanks Martin Varmus

function floorFigure(figure, decimals){
     if (!decimals) decimals = 2;
     var d = Math.pow(10,decimals);
     return ((figure*d)/d).toFixed(decimals);
};

floorFigure(123.5999)    =>   "123.59"
floorFigure(123.5999, 3) =>   "123.599"

I make a simple update and I got proper rounding. The update is following line

return ((figure*d)/d).toFixed(decimals);

remove parseInt() function

The following code works very good for me:

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

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;
}

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

Reference Answer

var number = 10.5859;_x000D_
var fixed2FloatPoints = parseInt(number * 100) / 100;_x000D_
console.log(fixed2FloatPoints);

Thank You !

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...

num = 19.66752
f = num.toFixed(3).slice(0,-1)
alert(f)

This will return 19.66

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;
}

Update (Jan 2021)

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) {
    var 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 {
    var 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)

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)

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.

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);

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;
}

Just truncate the digits:

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

Taking the knowledge from all the previous answers combined,

this is what I came up with as a solution:

function toFixedWithoutRounding(num, fractionDigits) {
  if ((num > 0 && num < 0.000001) || (num < 0 && num > -0.000001)) {
    // HACK: below this js starts to turn numbers into exponential form like 1e-7.
    // This gives wrong results so we are just changing the original number to 0 here
    // as we don't need such small numbers anyway.
    num = 0;
  }
  const re = new RegExp('^-?\\d+(?:\.\\d{0,' + (fractionDigits || -1) + '})?');
  return Number(num.toString().match(re)[0]).toFixed(fractionDigits);
}

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.

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;

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

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;
};

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

function toFixed(num, fixed) {
    fixed = fixed || 0;
    var front = Math.floor(num);
    var back = 0;
    for (var i = 1; i <= fixed; i++) {
        var value = Math.floor(num * Math.pow(10, i)) % 10;
        back += value / Math.pow(10, i);
    }
    return front + back;
}

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

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)

I fixed using following simple way-

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

Results will be 15.77

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

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;

Here is another one variant to save .toFixed([digits]) functional without rounding float variable:

Number.prototype.toRealFixed = function(digits) {
    return Math.floor(this.valueOf() * Math.pow(10, digits)) / Math.pow(10, digits);
};

And calling:

var float_var = 0.02209062;
float_var.toRealFixed();

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!

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

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

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?

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.