I am validating a decimal number using JavaScript.
Am just using NaN
var a = 12345.67
Is there any javascript function to get the count or the value itself before and after decimal point .
before() should return 1234
after() should return 67
Please dont suggest a substring!
var a = 12345.67;
alert(a.toString().split(".")[0]); ///before
alert(a.toString().split(".")[1]); ///after
Here is a simple fiddle http://jsfiddle.net/qWtSc/
zzzzBov's suggestion is this
Number.prototype.before = function () {
var value = parseInt(this.toString().split(".")[0], 10);//before
return value ? value : 0;
}
Number.prototype.after = function () {
var value = parseInt(this.toString().split(".")[1], 10);//after
return value ? value : 0;
}
Usage
alert(a.before()); ///before
alert(a.after()); ///after
before is easy. It's just a round down operation.
var before = function(n) {
return Math.floor(n);
};
after is harder without string processing. I mean how would you handle after(Math.PI)? You can't hold a integer with an infinite number of digits after all.
But with some string processing it's easy, just know it won't be exact due to the wonders of floating point math.
var after = function(n) {
var fraction = n.toString().split('.')[1];
return parseInt(fraction, 10);
};
Playing off of other answers... and you wanted a 'numeric' version.. still easiest to convert it to a string and work off the split function...
function getNatural(num) {
return parseFloat(num.toString().split(".")[0]);
}
function getDecimal(num) {
return parseFloat(num.toString().split(".")[1]);
}
var a = 12345.67;
alert(getNatural(a)); ///before
alert(getDecimal(a)); ///after
http://jsfiddle.net/rlemon/qWtSc/1/
var decimalPlaces = 2;
var num = 12345.673
var roundedDecimal = num.toFixed(decimalPlaces);
var intPart = Math.floor(roundedDecimal);
var fracPart = parseInt((roundedDecimal - intPart), 10);
//or
var fractPart = (roundedDecimal - intPart) * Math.pow(10, decimalPlaces);
To find the count/length of characters after dot:
var a = 12345.67;
var after_dot = (a.toString().split(".")[1]).length;
var before_dot= (a.toString().split(".")[0]).length;
Unfortunately there's no way to get the factional part in a reliable way using math functions, because pretty odd roundings often occur, depending on the Javascript engine used.
The best thing to do is to convert it to a string, and then checking if the results is in decimal or scientific notation.
Number.prototype.after = function() {
var string = this.toString();
var epos = string.indexOf("e");
if (epos === -1) { // Decimal notation
var i = string.indexOf(".");
return i === -1 ? "" : n.substring(i + 1);
}
// Scientific notation
var exp = string.substring(epos + 1) - 0; // this is actually faster
// than parseInt in many browsers
var mantix = n.string.substring(0, epos).replace(".", "");
if (exp >= -1) return mantix.substring(exp + 1);
for (; exp < -1; exp++) mantix = "0" + mantix;
return mantix;
}
If your digits after decimal point are fixed, then this solution works without converting to string.
This example shows a solution for 2 digits after decimal.
Before decimal:
const wholeNum = Math.floor(num);
After decimal:
let decimal = (num - wholeNum) * 100
Goal
I am at the final stage of scripting a Luhn algorithm.
Problem
Let's say I have a final calculation of 73
How can I round it up to the next 0? So the final value is 80.
And lastly, how can I get the value that made the addition? e.g. 7 is the final answer.
Current code
function validateCred(array) {
// Array to return the result of the algorithm
const algorithmValue = [];
// Create a [2, 1, 2] Pattern
const pattern = array.map((x, y) => {
return 2 - (y % 2);
});
// From given array, multiply each element by it's pattern
const multiplyByPattern = array.map((n, i) => {
return n * pattern[i];
});
// From the new array, split the numbers with length of 2 e.g. 12 and add them together e.g. 1 + 2 = 3
multiplyByPattern.forEach(el => {
// Check for lenght of 2
if(el.toString().length == 2) {
// Split the number
const splitNum = el.toString().split('');
// Add the 2 numbers together
const addSplitNum = splitNum.map(Number).reduce(add, 0);
// Function to add number together
function add(accumalator, a) {
return accumalator + a;
}
algorithmValue.push(addSplitNum);
}
// Check for lenght of 1
else if(el.toString().length == 1){
algorithmValue.push(el);
}
});
// Sum up the algorithmValue together
const additionOfAlgorithmValue = algorithmValue.reduce((a, b) => {
return a + b;
});
// Mod the final value by 10
if((additionOfAlgorithmValue % 10) == 0) {
return true;
}
else{
return false;
}
}
// Output is False
console.log(validateCred([2,7,6,9,1,4,8,3,0,4,0,5,9,9,8]));
Summary of the code above
The output should be True. This is because, I have given the total length of 15 digits in the array. Whereas it should be 16. I know the 16th value is 7, because the total value of the array given is 73, and rounding it up to the next 0 is 80, meaning the check digit is 7.
Question
How can I get the check number if given array length is less than 15?
You could do something like this:
let x = [73,81,92,101,423];
let y = x.map((v) => {
let remainder = v % 10;
let nextRounded = v + (10-remainder);
/* or you could use
let nextRounded = (parseInt(v/10)+1)*10;
*/
let amountToNextRounded = 10 - remainder;
return [nextRounded,amountToNextRounded];
});
console.log(y);
EDIT
As noticed by #pilchard you could find nextRounded using this more simplified way:
let nextRounded = v + (10-remainder);
https://stackoverflow.com/users/13762301/pilchard
I think what you need is this:
var oldNum = 73
var newNum = Math.ceil((oldNum+1) / 10) * 10;;
Then check the difference using this:
Math.abs(newNum - oldNum);
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 am validating a decimal number using JavaScript.
Am just using NaN
var a = 12345.67
Is there any javascript function to get the count or the value itself before and after decimal point .
before() should return 1234
after() should return 67
Please dont suggest a substring!
var a = 12345.67;
alert(a.toString().split(".")[0]); ///before
alert(a.toString().split(".")[1]); ///after
Here is a simple fiddle http://jsfiddle.net/qWtSc/
zzzzBov's suggestion is this
Number.prototype.before = function () {
var value = parseInt(this.toString().split(".")[0], 10);//before
return value ? value : 0;
}
Number.prototype.after = function () {
var value = parseInt(this.toString().split(".")[1], 10);//after
return value ? value : 0;
}
Usage
alert(a.before()); ///before
alert(a.after()); ///after
before is easy. It's just a round down operation.
var before = function(n) {
return Math.floor(n);
};
after is harder without string processing. I mean how would you handle after(Math.PI)? You can't hold a integer with an infinite number of digits after all.
But with some string processing it's easy, just know it won't be exact due to the wonders of floating point math.
var after = function(n) {
var fraction = n.toString().split('.')[1];
return parseInt(fraction, 10);
};
Playing off of other answers... and you wanted a 'numeric' version.. still easiest to convert it to a string and work off the split function...
function getNatural(num) {
return parseFloat(num.toString().split(".")[0]);
}
function getDecimal(num) {
return parseFloat(num.toString().split(".")[1]);
}
var a = 12345.67;
alert(getNatural(a)); ///before
alert(getDecimal(a)); ///after
http://jsfiddle.net/rlemon/qWtSc/1/
var decimalPlaces = 2;
var num = 12345.673
var roundedDecimal = num.toFixed(decimalPlaces);
var intPart = Math.floor(roundedDecimal);
var fracPart = parseInt((roundedDecimal - intPart), 10);
//or
var fractPart = (roundedDecimal - intPart) * Math.pow(10, decimalPlaces);
To find the count/length of characters after dot:
var a = 12345.67;
var after_dot = (a.toString().split(".")[1]).length;
var before_dot= (a.toString().split(".")[0]).length;
Unfortunately there's no way to get the factional part in a reliable way using math functions, because pretty odd roundings often occur, depending on the Javascript engine used.
The best thing to do is to convert it to a string, and then checking if the results is in decimal or scientific notation.
Number.prototype.after = function() {
var string = this.toString();
var epos = string.indexOf("e");
if (epos === -1) { // Decimal notation
var i = string.indexOf(".");
return i === -1 ? "" : n.substring(i + 1);
}
// Scientific notation
var exp = string.substring(epos + 1) - 0; // this is actually faster
// than parseInt in many browsers
var mantix = n.string.substring(0, epos).replace(".", "");
if (exp >= -1) return mantix.substring(exp + 1);
for (; exp < -1; exp++) mantix = "0" + mantix;
return mantix;
}
If your digits after decimal point are fixed, then this solution works without converting to string.
This example shows a solution for 2 digits after decimal.
Before decimal:
const wholeNum = Math.floor(num);
After decimal:
let decimal = (num - wholeNum) * 100