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Show next date while skipping holidays in javascript [duplicate]

I have a javascript function which is calculating working days between 2 dates, it works, but the problem is that it not consider holidays. How can I modify this function, for example by adding holidays in exception array?
Searched in internet about this question, but haven't find about holidays exception.
For example holidays array:
var holidays = ['2016-05-03','2016-05-05'];
And I have a functions to calculate this:
function workingDaysBetweenDates(d0, d1) {
var startDate = parseDate(d0);
var endDate = parseDate(d1);
// Validate input
if (endDate < startDate)
return 0;
// Calculate days between dates
var millisecondsPerDay = 86400 * 1000; // Day in milliseconds
startDate.setHours(0,0,0,1); // Start just after midnight
endDate.setHours(23,59,59,999); // End just before midnight
var diff = endDate - startDate; // Milliseconds between datetime objects
var days = Math.ceil(diff / millisecondsPerDay);
// Subtract two weekend days for every week in between
var weeks = Math.floor(days / 7);
days = days - (weeks * 2);
// Handle special cases
var startDay = startDate.getDay();
var endDay = endDate.getDay();
// Remove weekend not previously removed.
if (startDay - endDay > 1)
days = days - 2;
// Remove start day if span starts on Sunday but ends before Saturday
if (startDay == 0 && endDay != 6)
days = days - 1
// Remove end day if span ends on Saturday but starts after Sunday
if (endDay == 6 && startDay != 0)
days = days - 1
return days;
}
function parseDate(input) {
// Transform date from text to date
var parts = input.match(/(\d+)/g);
// new Date(year, month [, date [, hours[, minutes[, seconds[, ms]]]]])
return new Date(parts[0], parts[1]-1, parts[2]); // months are 0-based
}
Have made an example in jsfiddle:
JSFiddle example
Maybe there are some other functions which can easy use in Jquery?

Try:
var startDate = new Date('05/03/2016');
var endDate = new Date('05/10/2016');
var numOfDates = getBusinessDatesCount(startDate,endDate);
function getBusinessDatesCount(startDate, endDate) {
let count = 0;
const curDate = new Date(startDate.getTime());
while (curDate <= endDate) {
const dayOfWeek = curDate.getDay();
if(dayOfWeek !== 0 && dayOfWeek !== 6) count++;
curDate.setDate(curDate.getDate() + 1);
}
alert(count);
return count;
}

The easiest way to achieve it is looking for these days between your begin and end date.
Edit: I added an additional verification to make sure that only working days from holidays array are subtracted.
$(document).ready(() => {
$('#calc').click(() => {
var d1 = $('#d1').val();
var d2 = $('#d2').val();
$('#dif').text(workingDaysBetweenDates(d1,d2));
});
});
let workingDaysBetweenDates = (d0, d1) => {
/* Two working days and an sunday (not working day) */
var holidays = ['2016-05-03', '2016-05-05', '2016-05-07'];
var startDate = parseDate(d0);
var endDate = parseDate(d1);
// Validate input
if (endDate <= startDate) {
return 0;
}
// Calculate days between dates
var millisecondsPerDay = 86400 * 1000; // Day in milliseconds
startDate.setHours(0, 0, 0, 1); // Start just after midnight
endDate.setHours(23, 59, 59, 999); // End just before midnight
var diff = endDate - startDate; // Milliseconds between datetime objects
var days = Math.ceil(diff / millisecondsPerDay);
// Subtract two weekend days for every week in between
var weeks = Math.floor(days / 7);
days -= weeks * 2;
// Handle special cases
var startDay = startDate.getDay();
var endDay = endDate.getDay();
// Remove weekend not previously removed.
if (startDay - endDay > 1) {
days -= 2;
}
// Remove start day if span starts on Sunday but ends before Saturday
if (startDay == 0 && endDay != 6) {
days--;
}
// Remove end day if span ends on Saturday but starts after Sunday
if (endDay == 6 && startDay != 0) {
days--;
}
/* Here is the code */
holidays.forEach(day => {
if ((day >= d0) && (day <= d1)) {
/* If it is not saturday (6) or sunday (0), substract it */
if ((parseDate(day).getDay() % 6) != 0) {
days--;
}
}
});
return days;
}
function parseDate(input) {
// Transform date from text to date
var parts = input.match(/(\d+)/g);
// new Date(year, month [, date [, hours[, minutes[, seconds[, ms]]]]])
return new Date(parts[0], parts[1]-1, parts[2]); // months are 0-based
}
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<input type="text" id="d1" value="2016-05-02"><br>
<input type="text" id="d2" value="2016-05-08">
<p>Working days count: <span id="dif"></span></p>
<button id="calc">Calc</button>
<p>
Now it shows 5 days, but I need for example add holidays
3 and 5 May (2016-05-03 and 2016-05-05) so the result will be 3 working days
</p>

I took a similar approach to #OscarGarcia mainly as an excercise since my JS is rusty.
While it looks similar, it takes care not to substract a day twice if a holiday happens to be on a saturday or sunday. This way, you can pre-load a list of recurring dates (such as Dec 25th, Jan 1st, July 4th, which may or may not be on an otherwise working day -monday thru friday-)
$(document).ready(function(){
$('#calc').click(function(){
var d1 = $('#d1').val();
var d2 = $('#d2').val();
$('#dif').text(workingDaysBetweenDates(d1,d2));
});
});
function workingDaysBetweenDates(d0, d1) {
var startDate = parseDate(d0);
var endDate = parseDate(d1);
// populate the holidays array with all required dates without first taking care of what day of the week they happen
var holidays = ['2018-12-09', '2018-12-10', '2018-12-24', '2018-12-31'];
// Validate input
if (endDate < startDate)
return 0;
var z = 0; // number of days to substract at the very end
for (i = 0; i < holidays.length; i++)
{
var cand = parseDate(holidays[i]);
var candDay = cand.getDay();
if (cand >= startDate && cand <= endDate && candDay != 0 && candDay != 6)
{
// we'll only substract the date if it is between the start or end dates AND it isn't already a saturday or sunday
z++;
}
}
// Calculate days between dates
var millisecondsPerDay = 86400 * 1000; // Day in milliseconds
startDate.setHours(0,0,0,1); // Start just after midnight
endDate.setHours(23,59,59,999); // End just before midnight
var diff = endDate - startDate; // Milliseconds between datetime objects
var days = Math.ceil(diff / millisecondsPerDay);
// Subtract two weekend days for every week in between
var weeks = Math.floor(days / 7);
days = days - (weeks * 2);
// Handle special cases
var startDay = startDate.getDay();
var endDay = endDate.getDay();
// Remove weekend not previously removed.
if (startDay - endDay > 1)
days = days - 2;
// Remove start day if span starts on Sunday but ends before Saturday
if (startDay == 0 && endDay != 6)
days = days - 1
// Remove end day if span ends on Saturday but starts after Sunday
if (endDay == 6 && startDay != 0)
days = days - 1
// substract the holiday dates from the original calculation and return to the DOM
return days - z;
}
function parseDate(input) {
// Transform date from text to date
var parts = input.match(/(\d+)/g);
// new Date(year, month [, date [, hours[, minutes[, seconds[, ms]]]]])
return new Date(parts[0], parts[1]-1, parts[2]); // months are 0-based
}
2018-12-09 is a sunday... with this code, it'll only be substracted once (for being a sunday) and not twice (as it would if we only checked if its a national holiday)

I think this solution is much more simpler
const numberOfDaysInclusive = (d0, d1) => {
return 1 + Math.round((d1.getTime()-d0.getTime())/(24*3600*1000));
}
const numberOfWeekends = (d0, d1) => {
const days = numberOfDaysInclusive(d0, d1); // total number of days
const sundays = Math.floor((days + (d0.getDay() + 6) % 7) / 7); // number of sundays
return 2*sundays + (d1.getDay()==6) - (d0.getDay()==0); // multiply sundays by 2 to get both sat and sun, +1 if d1 is saturday, -1 if d0 is sunday
}
const numberOfWeekdays = (d0, d1) => {
return numberOfDaysInclusive(d0, d1) - numberOfWeekends(d0, d1);
}

Get all weekdays between two dates:
private getCorrectWeekDays(StartDate,EndDate){
let _weekdays = [0,1,2,3,4];
var wdArr= [];
var currentDate = StartDate;
while (currentDate <= EndDate) {
if ( _weekdays.includes(currentDate.getDay())){
wdArr.push(currentDate);
//if you want to format it to yyyy-mm-dd
//wdArr.push(currentDate.toISOString().split('T')[0]);
}
currentDate.setDate(currentDate.getDate() +1);
}
return wdArr;
}

You can also try this piece of code:
const moment = require('moment-business-days');
/**
*
* #param {String} date - iso Date
* #returns {Number} difference between now and #param date
*/
const calculateDaysLeft = date => {
try {
return moment(date).businessDiff(moment(new Date()))
} catch (err) {
throw new Error(err)
}
}

The top answer actually works but with a flaw.
When the holyday is in a Saturday or Sunday it still reduces a day.
Add this to the existing code:
.... /* Here is the code */
for (var i in holidays) {
if ((holidays[i] >= d0) && (holidays[i] <= d1)) {
// Check if specific holyday is Saturday or Sunday
var yourDate = new Date(holidays[i]);
if(yourDate.getDay() === 6 || yourDate.getDay() === 0){
// If it is.. do nothing
} else {
// if it is not, reduce a day..
days--;
}
}
}

const workday_count = (start, end) => {
start = moment(start).format(("YYYY-MM-DD"))
end = moment(end).format(("YYYY-MM-DD"))
let workday_count = 0;
let totalDays = moment(end).diff(moment(start), "days");
let date = start
for (let i = 1; i <= totalDays; i++) {
if (i == 1) {
date = moment(date)
} else {
date = moment(date).add(1, "d");
}
date = new Date(date);
let dayOfWeek = date.getDay();
let isWeekend = (dayOfWeek === 6) || (dayOfWeek === 0);
if (!isWeekend) {
workday_count = workday_count + 1;
}
}
return workday_count;
}

Simply reduce the length of array from the value you have got (in your fiddle)
var numberofdayswithoutHolidays= 5;
var holidays = ['2016-05-03','2016-05-05'];
alert( numberofdayswithoutHolidays - holidays.length );
You need to filter out weekends from holidays as well
holidays = holidays.filter( function(day){
var day = parseDate( day ).getDay();
return day > 0 && day < 6;
})

$(document).ready(() => {
$('#calc').click(() => {
var d1 = $('#d1').val();
var d2 = $('#d2').val();
$('#dif').text(workingDaysBetweenDates(d1,d2));
});
});
let workingDaysBetweenDates = (d0, d1) => {
/* Two working days and an sunday (not working day) */
var holidays = ['2016-05-03', '2016-05-05', '2016-05-07'];
var startDate = parseDate(d0);
var endDate = parseDate(d1);
// Validate input
if (endDate < startDate) {
return 0;
}
// Calculate days between dates
var millisecondsPerDay = 86400 * 1000; // Day in milliseconds
startDate.setHours(0, 0, 0, 1); // Start just after midnight
endDate.setHours(23, 59, 59, 999); // End just before midnight
var diff = endDate - startDate; // Milliseconds between datetime objects
var days = Math.ceil(diff / millisecondsPerDay);
// Subtract two weekend days for every week in between
var weeks = Math.floor(days / 7);
days -= weeks * 2;
// Handle special cases
var startDay = startDate.getDay();
var endDay = endDate.getDay();
// Remove weekend not previously removed.
if (startDay - endDay > 1) {
days -= 2;
}
// Remove start day if span starts on Sunday but ends before Saturday
if (startDay == 0 && endDay != 6) {
days--;
}
// Remove end day if span ends on Saturday but starts after Sunday
if (endDay == 6 && startDay != 0) {
days--;
}
/* Here is the code */
holidays.forEach(day => {
if ((day >= d0) && (day <= d1)) {
/* If it is not saturday (6) or sunday (0), substract it */
if ((parseDate(day).getDay() % 6) != 0) {
days--;
}
}
});
return days;
}
function parseDate(input) {
// Transform date from text to date
var parts = input.match(/(\d+)/g);
// new Date(year, month [, date [, hours[, minutes[, seconds[, ms]]]]])
return new Date(parts[0], parts[1]-1, parts[2]); // months are 0-based
}
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<input type="text" id="d1" value="2016-05-02"><br>
<input type="text" id="d2" value="2016-05-08">
<p>Working days count: <span id="dif"></span></p>
<button id="calc">Calc</button>
<p>
Now it shows 5 days, but I need for example add holidays
3 and 5 May (2016-05-03 and 2016-05-05) so the result will be 3 working days
</p>

how can i count the range between 2 dates without counting the weekend in javascript

i want to calculate the range between 2 dates without counting weekend in javascript. i have some code that already count the range between them. but i'm stuck with the weekend part. date inputed by CJuiDatePicker in YII framework
<script>
function calcDay(dt1, dt2, range){
var msec1 = dt1;
var date1 = new date(msec1);
var msec2 = dt2;
var date2 = new date(msec2);
if(date1>0 || date2>0){
range.val(isFinite(Math.round(date2-date1)/86400000) || 0);
}
};
</script>
86400000 is day in millisecond
thanks in advance

The function you'll need is getUTCDay().
the pseudo code would be as follows:
1 - determine full weeks (days/7 truncated)
2 - calculate weekday/weekend: 2 * result = weekend days, 5 * result = weekdays.
3 - after that, remainder and starting day of week will determine the 1 or 2 additional days
Hope that helps, let me know if you need the javascript,
- John
Edited, as requested. NOTE: tweaked your original for testing, you should spot the needed changes to restore.
function calcDay(dt1, dt2, range)
{
var msec1 = "October 13, 2014 11:13:00";
var date1 = new Date(msec1);
var msec2 = "October 13, 2013 11:13:00";
var date2 = new Date(msec2);
var days;
var wdays;
var startday;
var nLeft;
// neither should be zero
if(date1>0 && date2>0) {
days = Math.round( Math.abs((date2-date1)/86400000) );
wdays = Math.round(days / 7) * 5;
nLeft = days % 7;
startday = (date1 > date2) ? date2.getUTCDay() : date1.getUTCDay();
if (startday < 2) {
wdays += Math.max(nLeft+startday-1,0);
} else if (startday == 6) {
wdays += Math.max(nLeft-2,0);
} else if (nLeft > (7-startday)) {
wdays += (nLeft-2)
} else {
wdays += Math.min(nLeft, 6-startday)
}
}
};

i found my own solution, but i forgot to share it. this is how i make it
function myUpdate(dt1, dt2,range){
var msec1 = dt1;
var date1 = new Date(msec1);
var msec2 = dt2;
var date2 = new Date(msec2);
var diff = (isFinite(Math.round (date2 - date1) / 86400000) && Math.round (date2 - date1) / 86400000 || 0);
var wEnd=0;
if(date1>0 || date2>0){
for(var i=0; i<=diff; i++){
if(date1.getDay() ==6 || date1.getDay()==0){
wEnd = wEnd + 1;
}
date1.setDate(date1.getDate() + 1);
}
}
range.val(Math.round((diff-wEnd)+1));
};
first u should count the different date, then the date1 will be check if it is sunday or saturday. then date1 will be added 1 till the value of date1 is equal to date2. if date1 is/are saturday or sunday, wEnd will gain 1. so u can substract diff with wEnd. hope this can help u guys

How can I calculate the number of years between two dates?

I want to get the number of years between two dates. I can get the number of days between these two days, but if I divide it by 365 the result is incorrect because some years have 366 days.
This is my code to get date difference:
var birthday = value;//format 01/02/1900
var dateParts = birthday.split("/");
var checkindate = new Date(dateParts[2], dateParts[0] - 1, dateParts[1]);
var now = new Date();
var difference = now - checkindate;
var days = difference / (1000*60*60*24);
var thisyear = new Date().getFullYear();
var birthyear = dateParts[2];
var number_of_long_years = 0;
for(var y=birthyear; y <= thisyear; y++){
if( (y % 4 == 0 && y % 100 == 0) || y % 400 == 0 ) {
number_of_long_years++;
}
}
The day count works perfectly. I am trying to do add the additional days when it is a 366-day year, and I'm doing something like this:
var years = ((days)*(thisyear-birthyear))
/((number_of_long_years*366) + ((thisyear-birthyear-number_of_long_years)*365) );
I'm getting the year count. Is this correct, or is there a better way to do this?

Sleek foundation javascript function.
function calculateAge(birthday) { // birthday is a date
var ageDifMs = Date.now() - birthday;
var ageDate = new Date(ageDifMs); // miliseconds from epoch
return Math.abs(ageDate.getUTCFullYear() - 1970);
}

Probably not the answer you're looking for, but at 2.6kb, I would not try to reinvent the wheel and I'd use something like moment.js. Does not have any dependencies.
The diff method is probably what you want: http://momentjs.com/docs/#/displaying/difference/

Using pure javascript Date(), we can calculate the numbers of years like below
document.getElementById('getYearsBtn').addEventListener('click', function () {
var enteredDate = document.getElementById('sampleDate').value;
// Below one is the single line logic to calculate the no. of years...
var years = new Date(new Date() - new Date(enteredDate)).getFullYear() - 1970;
console.log(years);
});
<input type="text" id="sampleDate" value="1980/01/01">
<div>Format: yyyy-mm-dd or yyyy/mm/dd</div><br>
<button id="getYearsBtn">Calculate Years</button>

No for-each loop, no extra jQuery plugin needed... Just call the below function.. Got from Difference between two dates in years
function dateDiffInYears(dateold, datenew) {
var ynew = datenew.getFullYear();
var mnew = datenew.getMonth();
var dnew = datenew.getDate();
var yold = dateold.getFullYear();
var mold = dateold.getMonth();
var dold = dateold.getDate();
var diff = ynew - yold;
if (mold > mnew) diff--;
else {
if (mold == mnew) {
if (dold > dnew) diff--;
}
}
return diff;
}

I use the following for age calculation.
I named it gregorianAge() because this calculation gives exactly how we denote age using Gregorian calendar. i.e. Not counting the end year if month and day is before the month and day of the birth year.
/**
* Calculates human age in years given a birth day. Optionally ageAtDate
* can be provided to calculate age at a specific date
*
* #param string|Date Object birthDate
* #param string|Date Object ageAtDate optional
* #returns integer Age between birthday and a given date or today
*/
gregorianAge = function(birthDate, ageAtDate) {
// convert birthDate to date object if already not
if (Object.prototype.toString.call(birthDate) !== '[object Date]')
birthDate = new Date(birthDate);
// use today's date if ageAtDate is not provided
if (typeof ageAtDate == "undefined")
ageAtDate = new Date();
// convert ageAtDate to date object if already not
else if (Object.prototype.toString.call(ageAtDate) !== '[object Date]')
ageAtDate = new Date(ageAtDate);
// if conversion to date object fails return null
if (ageAtDate == null || birthDate == null)
return null;
var _m = ageAtDate.getMonth() - birthDate.getMonth();
// answer: ageAt year minus birth year less one (1) if month and day of
// ageAt year is before month and day of birth year
return (ageAtDate.getFullYear()) - birthDate.getFullYear()
- ((_m < 0 || (_m === 0 && ageAtDate.getDate() < birthDate.getDate()))?1:0)
}
<input type="text" id="birthDate" value="12 February 1982">
<div style="font-size: small; color: grey">Enter a date in an acceptable format e.g. 10 Dec 2001</div><br>
<button onClick='js:alert(gregorianAge(document.getElementById("birthDate").value))'>What's my age?</button>

Little out of date but here is a function you can use!
function calculateAge(birthMonth, birthDay, birthYear) {
var currentDate = new Date();
var currentYear = currentDate.getFullYear();
var currentMonth = currentDate.getMonth();
var currentDay = currentDate.getDate();
var calculatedAge = currentYear - birthYear;
if (currentMonth < birthMonth - 1) {
calculatedAge--;
}
if (birthMonth - 1 == currentMonth && currentDay < birthDay) {
calculatedAge--;
}
return calculatedAge;
}
var age = calculateAge(12, 8, 1993);
alert(age);

You can get the exact age using timesstamp:
const getAge = (dateOfBirth, dateToCalculate = new Date()) => {
const dob = new Date(dateOfBirth).getTime();
const dateToCompare = new Date(dateToCalculate).getTime();
const age = (dateToCompare - dob) / (365 * 24 * 60 * 60 * 1000);
return Math.floor(age);
};

let currentTime = new Date().getTime();
let birthDateTime= new Date(birthDate).getTime();
let difference = (currentTime - birthDateTime)
var ageInYears=difference/(1000*60*60*24*365)

Yep, moment.js is pretty good for this:
var moment = require('moment');
var startDate = new Date();
var endDate = new Date();
endDate.setDate(endDate.getFullYear() + 5); // Add 5 years to second date
console.log(moment.duration(endDate - startDate).years()); // This should returns 5

getYears(date1, date2) {
let years = new Date(date1).getFullYear() - new Date(date2).getFullYear();
let month = new Date(date1).getMonth() - new Date(date2).getMonth();
let dateDiff = new Date(date1).getDay() - new Date(date2).getDay();
if (dateDiff < 0) {
month -= 1;
}
if (month < 0) {
years -= 1;
}
return years;
}

for(var y=birthyear; y <= thisyear; y++){
if( (y % 4 == 0 && y % 100 == 0) || y % 400 == 0 ) {
days = days-366;
number_of_long_years++;
} else {
days=days-365;
}
year++;
}
can you try this way??

function getYearDiff(startDate, endDate) {
let yearDiff = endDate.getFullYear() - startDate.getFullYear();
if (startDate.getMonth() > endDate.getMonth()) {
yearDiff--;
} else if (startDate.getMonth() === endDate.getMonth()) {
if (startDate.getDate() > endDate.getDate()) {
yearDiff--;
} else if (startDate.getDate() === endDate.getDate()) {
if (startDate.getHours() > endDate.getHours()) {
yearDiff--;
} else if (startDate.getHours() === endDate.getHours()) {
if (startDate.getMinutes() > endDate.getMinutes()) {
yearDiff--;
}
}
}
}
return yearDiff;
}
alert(getYearDiff(firstDate, secondDate));

getAge(month, day, year) {
let yearNow = new Date().getFullYear();
let monthNow = new Date().getMonth() + 1;
let dayNow = new Date().getDate();
if (monthNow === month && dayNow < day || monthNow < month) {
return yearNow - year - 1;
} else {
return yearNow - year;
}
}

If you are using moment
/**
* Convert date of birth into age
* param {string} dateOfBirth - date of birth
* param {string} dateToCalculate - date to compare
* returns {number} - age
*/
function getAge(dateOfBirth, dateToCalculate) {
const dob = moment(dateOfBirth);
return moment(dateToCalculate).diff(dob, 'years');
};

If you want to calculate the years and keep the remainder of the time left for further calculations you can use this function most of the other answers discard the remaining time.
It returns the years and the remainder in milliseconds. This is useful if you want to calculate the time (days or minutes) left after you calculate the years.
The function works by first calculating the difference in years directly using *date.getFullYear()*.
Then it checks if the last year between the two dates is up to a full year by setting the two dates to the same year.
Eg:
oldDate= 1 July 2020,
newDate= 1 June 2022,
years =2020 -2022 =2
Now set old date to new date's year 2022
oldDate = 1 July, 2022
If the last year is not up to a full year then the year is subtracted by 1, the old date is set to the previous year and the interval from the previous year to the current date is calculated to give the remainder in milliseconds.
In the example since old date July 2022 is greater than June 2022 then it means a full year has not yet elapsed (from July 2021 to June 2022) therefore the year count is greater by 1. So years should be decreased by 1. And the actual year count from July 2020 to June 2022 is 1 year ,... months.
If the last year is a full year then the year count by *date.getFullYear()* is correct and the time that has elapsed from the current old date to new date is calculated as the remainder.
If old date= 1 April, 2020, new date = 1 June, 2022 and old date is set to April 2022 after calculating the year =2.
Eg: from April 2020 to June 2022 a duration of 2 years has passed with the remainder being the time from April 2022 to June 2022.
There are also checks for cases where the two dates are in the same year and if the user enters the dates in the wrong order the new Date is less recent than the old Date.
let getYearsAndRemainder = (newDate, oldDate) => {
let remainder = 0;
// get initial years between dates
let years = newDate.getFullYear() - oldDate.getFullYear();
if (years < 0) {// check to make sure the oldDate is the older of the two dates
console.warn('new date is lesser than old date in year difference')
years = 0;
} else {
// set the old date to the same year as new date
oldDate.setFullYear(newDate.getFullYear());
// check if the old date is less than new date in the same year
if (oldDate - newDate > 0) {
//if true, the old date is greater than the new date
// the last but one year between the two dates is not up to a year
if (years != 0) {// dates given in inputs are in the same year, no need to calculate years if the number of years is 0
console.log('Subtracting year');
//set the old year to the previous year
years--;
oldDate.setFullYear(oldDate.getFullYear() - 1);
}
}
}
//calculate the time difference between the old year and newDate.
remainder = newDate - oldDate;
if (remainder < 0) { //check for negative dates due to wrong inputs
console.warn('old date is greater than new Date');
console.log('new date', newDate, 'old date', oldDate);
}
return { years, remainder };
}
let old = new Date('2020-07-01');
console.log( getYearsAndRemainder(new Date(), old));

Date calculation work via the Julian day number. You have to take the first of January of the two years. Then you convert the Gregorian dates into Julian day numbers and after that you take just the difference.

Maybe my function can explain better how to do this in a simple way without loop, calculations and/or libs
function checkYearsDifference(birthDayDate){
var todayDate = new Date();
var thisMonth = todayDate.getMonth();
var thisYear = todayDate.getFullYear();
var thisDay = todayDate.getDate();
var monthBirthday = birthDayDate.getMonth();
var yearBirthday = birthDayDate.getFullYear();
var dayBirthday = birthDayDate.getDate();
//first just make the difference between years
var yearDifference = thisYear - yearBirthday;
//then check months
if (thisMonth == monthBirthday){
//if months are the same then check days
if (thisDay<dayBirthday){
//if today day is before birthday day
//then I have to remove 1 year
//(no birthday yet)
yearDifference = yearDifference -1;
}
//if not no action because year difference is ok
}
else {
if (thisMonth < monthBirthday) {
//if actual month is before birthday one
//then I have to remove 1 year
yearDifference = yearDifference -1;
}
//if not no action because year difference is ok
}
return yearDifference;
}

Bro, moment.js is awesome for this:
The diff method is what you want: http://momentjs.com/docs/#/displaying/difference/
The below function return array of years from the year to the current year.
const getYears = (from = 2017) => {
const diff = moment(new Date()).diff(new Date(`01/01/${from}`), 'years') ;
return [...Array(diff >= 0 ? diff + 1 : 0).keys()].map((num) => {
return from + num;
});
}
console.log(getYears(2016));
<script src="https://momentjs.com/downloads/moment.js"></script>

function dateDiffYearsOnly( dateNew,dateOld) {
function date2ymd(d){ w=new Date(d);return [w.getFullYear(),w.getMonth(),w.getDate()]}
function ymd2N(y){return (((y[0]<<4)+y[1])<<5)+y[2]} // or 60 and 60 // or 13 and 32 // or 25 and 40 //// with ...
function date2N(d){ return ymd2N(date2ymd(d))}
return (date2N(dateNew)-date2N(dateOld))>>9
}
test:
dateDiffYearsOnly(Date.now(),new Date(Date.now()-7*366*24*3600*1000));
dateDiffYearsOnly(Date.now(),new Date(Date.now()-7*365*24*3600*1000))

I went for the following very simple solution. It does not assume you were born in 1970 and it also takes into account the hour of the given birthday date.
function age(birthday) {
let now = new Date();
let year = now.getFullYear();
let years = year - birthday.getFullYear();
birthday = new Date(birthday.getTime()); // clone
birthday.setFullYear(year);
return now >= birthday ? years : years - 1;
}

This one Help you...
$("[id$=btnSubmit]").click(function () {
debugger
var SDate = $("[id$=txtStartDate]").val().split('-');
var Smonth = SDate[0];
var Sday = SDate[1];
var Syear = SDate[2];
// alert(Syear); alert(Sday); alert(Smonth);
var EDate = $("[id$=txtEndDate]").val().split('-');
var Emonth = EDate[0];
var Eday = EDate[1];
var Eyear = EDate[2];
var y = parseInt(Eyear) - parseInt(Syear);
var m, d;
if ((parseInt(Emonth) - parseInt(Smonth)) > 0) {
m = parseInt(Emonth) - parseInt(Smonth);
}
else {
m = parseInt(Emonth) + 12 - parseInt(Smonth);
y = y - 1;
}
if ((parseInt(Eday) - parseInt(Sday)) > 0) {
d = parseInt(Eday) - parseInt(Sday);
}
else {
d = parseInt(Eday) + 30 - parseInt(Sday);
m = m - 1;
}
// alert(y + " " + m + " " + d);
$("[id$=lblAge]").text("your age is " + y + "years " + m + "month " + d + "days");
return false;
});

if someone needs for interest calculation year in float format
function floatYearDiff(olddate, newdate) {
var new_y = newdate.getFullYear();
var old_y = olddate.getFullYear();
var diff_y = new_y - old_y;
var start_year = new Date(olddate);
var end_year = new Date(olddate);
start_year.setFullYear(new_y);
end_year.setFullYear(new_y+1);
if (start_year > newdate) {
start_year.setFullYear(new_y-1);
end_year.setFullYear(new_y);
diff_y--;
}
var diff = diff_y + (newdate - start_year)/(end_year - start_year);
return diff;
}

How to get date between start date and end date? (excluding date on weekend)

Hi want to get a date list between start date and end date. For example start date is 27-08-2010 and end date is 31-08-2010. So the date list is 27-08-2010,30-08-2010 and 31-08-2010. 29-08-2010 and 30-08-2010 will be ignore because it is in the weekend. I attach the picture for more clearer explanation. How to achieve this using javascript or jquery? I only want to get the list of date, for the business day calculation already done.

First of all, we can disable in the datepickers the holidays and the weekends, for that we'll be setting it like this in each datepicker component:
Disable weekends and holidays from DatePicker
var disabledDays = ["10-22-2010","8-16-2010"];
$("#startDate").datepicker({
constrainInput: true,
beforeShowDay: noWeekendsOrHolidays
});
function nationalDays(date) {
var m = date.getMonth(), d = date.getDate(), y = date.getFullYear();
for (i = 0; i < disabledDays.length; i++) {
if($.inArray((m+1) + '-' + d + '-' + y,disabledDays) != -1 || new Date() > date) {
return [false];
}
}
return [true];
}
function noWeekendsOrHolidays(date) {
var noWeekend = jQuery.datepicker.noWeekends(date);
return noWeekend[0] ? nationalDays(date) : noWeekend;
}
Now that we have the behavior to not let users select weekends or holidays as startdate or enddates we continue, calculating the total days between those dates:
Calculate Business Dates between two dates
as Find day difference between two dates (excluding weekend days) we have this function to calculate business days between dates:
function calcBusinessDays(dDate1, dDate2) { // input given as Date objects
var iWeeks, iDateDiff, iAdjust = 0;
if (dDate2 < dDate1) return -1; // error code if dates transposed
var iWeekday1 = dDate1.getDay(); // day of week
var iWeekday2 = dDate2.getDay();
iWeekday1 = (iWeekday1 == 0) ? 7 : iWeekday1; // change Sunday from 0 to 7
iWeekday2 = (iWeekday2 == 0) ? 7 : iWeekday2;
if ((iWeekday1 > 5) && (iWeekday2 > 5)) iAdjust = 1; // adjustment if both days on weekend
iWeekday1 = (iWeekday1 > 5) ? 5 : iWeekday1; // only count weekdays
iWeekday2 = (iWeekday2 > 5) ? 5 : iWeekday2;
// calculate differnece in weeks (1000mS * 60sec * 60min * 24hrs * 7 days = 604800000)
iWeeks = Math.floor((dDate2.getTime() - dDate1.getTime()) / 604800000)
if (iWeekday1 <= iWeekday2) {
iDateDiff = (iWeeks * 5) + (iWeekday2 - iWeekday1)
} else {
iDateDiff = ((iWeeks + 1) * 5) - (iWeekday1 - iWeekday2)
}
iDateDiff -= iAdjust // take into account both days on weekend
return (iDateDiff + 1); // add 1 because dates are inclusive
}
Count down Holidays
I'm right now in a hurry, what you need to do now is --iDateDiff (just before the return) for every date in disabledDays that is between dDate1 and dDate2.
How will you accomplish that?... You will iterate over the disabledDays array and convert/parse it to the Date Object and evaluate like:
var holiDay = Date.parse(iteratedDate);
if((holiDay >= dDate1 && holiDay <= dDate2)) {
--iDateDiff
}
I hope i've helped you...
EDITED:
Sorry, misunderstood the question, try this:
var Weekday = new Array("Sun","Mon","Tue","Wed","Thuy","Fri","Sat");
while (startDate<=endDate)
{
var weekDay = startDate.getDay();
if (weekDay < 6 && weekDay > 0) {
var month = startDate.getMonth()+1;
if( month <= 9 ) { month = "0"+month; }
var day = startDate.getDate();
if( day <= 9 ) { day = "0"+day; }
document.write(day+"/"+month+"/"+startDate.getFullYear() + " ("+Weekday[weekDay]+")<br />");
}
startDate.setDate(startDate.getDate()+1)
}
Live DEMO

Difference in Months between two dates in JavaScript

How would I work out the difference for two Date() objects in JavaScript, while only return the number of months in the difference?
Any help would be great :)

The definition of "the number of months in the difference" is subject to a lot of interpretation. :-)
You can get the year, month, and day of month from a JavaScript date object. Depending on what information you're looking for, you can use those to figure out how many months are between two points in time.
For instance, off-the-cuff:
function monthDiff(d1, d2) {
var months;
months = (d2.getFullYear() - d1.getFullYear()) * 12;
months -= d1.getMonth();
months += d2.getMonth();
return months <= 0 ? 0 : months;
}
function monthDiff(d1, d2) {
var months;
months = (d2.getFullYear() - d1.getFullYear()) * 12;
months -= d1.getMonth();
months += d2.getMonth();
return months <= 0 ? 0 : months;
}
function test(d1, d2) {
var diff = monthDiff(d1, d2);
console.log(
d1.toISOString().substring(0, 10),
"to",
d2.toISOString().substring(0, 10),
":",
diff
);
}
test(
new Date(2008, 10, 4), // November 4th, 2008
new Date(2010, 2, 12) // March 12th, 2010
);
// Result: 16
test(
new Date(2010, 0, 1), // January 1st, 2010
new Date(2010, 2, 12) // March 12th, 2010
);
// Result: 2
test(
new Date(2010, 1, 1), // February 1st, 2010
new Date(2010, 2, 12) // March 12th, 2010
);
// Result: 1
(Note that month values in JavaScript start with 0 = January.)
Including fractional months in the above is much more complicated, because three days in a typical February is a larger fraction of that month (~10.714%) than three days in August (~9.677%), and of course even February is a moving target depending on whether it's a leap year.
There are also some date and time libraries available for JavaScript that probably make this sort of thing easier.
Note: There used to be a + 1 in the above, here:
months = (d2.getFullYear() - d1.getFullYear()) * 12;
months -= d1.getMonth() + 1;
// −−−−−−−−−−−−−−−−−−−−^^^^
months += d2.getMonth();
That's because originally I said:
...this finds out how many full months lie between two dates, not counting partial months (e.g., excluding the month each date is in).
I've removed it for two reasons:
Not counting partial months turns out not to be what many (most?) people coming to the answer want, so I thought I should separate them out.
It didn't always work even by that definition. :-D (Sorry.)

If you do not consider the day of the month, this is by far the simpler solution
function monthDiff(dateFrom, dateTo) {
return dateTo.getMonth() - dateFrom.getMonth() +
(12 * (dateTo.getFullYear() - dateFrom.getFullYear()))
}
//examples
console.log(monthDiff(new Date(2000, 01), new Date(2000, 02))) // 1
console.log(monthDiff(new Date(1999, 02), new Date(2000, 02))) // 12 full year
console.log(monthDiff(new Date(2009, 11), new Date(2010, 0))) // 1
Be aware that month index is 0-based. This means that January = 0 and December = 11.

Here's a function that accurately provides the number of months between 2 dates.
The default behavior only counts whole months, e.g. 3 months and 1 day will result in a difference of 3 months. You can prevent this by setting the roundUpFractionalMonths param as true, so a 3 month and 1 day difference will be returned as 4 months.
The accepted answer above (T.J. Crowder's answer) isn't accurate, it returns wrong values sometimes.
For example, monthDiff(new Date('Jul 01, 2015'), new Date('Aug 05, 2015')) returns 0 which is obviously wrong. The correct difference is either 1 whole month or 2 months rounded-up.
Here's the function I wrote:
function getMonthsBetween(date1,date2,roundUpFractionalMonths)
{
//Months will be calculated between start and end dates.
//Make sure start date is less than end date.
//But remember if the difference should be negative.
var startDate=date1;
var endDate=date2;
var inverse=false;
if(date1>date2)
{
startDate=date2;
endDate=date1;
inverse=true;
}
//Calculate the differences between the start and end dates
var yearsDifference=endDate.getFullYear()-startDate.getFullYear();
var monthsDifference=endDate.getMonth()-startDate.getMonth();
var daysDifference=endDate.getDate()-startDate.getDate();
var monthCorrection=0;
//If roundUpFractionalMonths is true, check if an extra month needs to be added from rounding up.
//The difference is done by ceiling (round up), e.g. 3 months and 1 day will be 4 months.
if(roundUpFractionalMonths===true && daysDifference>0)
{
monthCorrection=1;
}
//If the day difference between the 2 months is negative, the last month is not a whole month.
else if(roundUpFractionalMonths!==true && daysDifference<0)
{
monthCorrection=-1;
}
return (inverse?-1:1)*(yearsDifference*12+monthsDifference+monthCorrection);
};

Sometimes you may want to get just the quantity of the months between two dates totally ignoring the day part. So for instance, if you had two dates- 2013/06/21 and 2013/10/18- and you only cared about the 2013/06 and 2013/10 parts, here are the scenarios and possible solutions:
var date1=new Date(2013,5,21);//Remember, months are 0 based in JS
var date2=new Date(2013,9,18);
var year1=date1.getFullYear();
var year2=date2.getFullYear();
var month1=date1.getMonth();
var month2=date2.getMonth();
if(month1===0){ //Have to take into account
month1++;
month2++;
}
var numberOfMonths;
1.If you want just the number of the months between the two dates excluding both month1 and month2
numberOfMonths = (year2 - year1) * 12 + (month2 - month1) - 1;
2.If you want to include either of the months
numberOfMonths = (year2 - year1) * 12 + (month2 - month1);
3.If you want to include both of the months
numberOfMonths = (year2 - year1) * 12 + (month2 - month1) + 1;

If you need to count full months, regardless of the month being 28, 29, 30 or 31 days. Below should work.
var months = to.getMonth() - from.getMonth()
+ (12 * (to.getFullYear() - from.getFullYear()));
if(to.getDate() < from.getDate()){
months--;
}
return months;
This is an extended version of the answer https://stackoverflow.com/a/4312956/1987208 but fixes the case where it calculates 1 month for the case from 31st of January to 1st of February (1day).
This will cover the following;
1st Jan to 31st Jan ---> 30days ---> will result in 0 (logical since it is not a full month)
1st Feb to 1st Mar ---> 28 or 29 days ---> will result in 1 (logical since it is a full month)
15th Feb to 15th Mar ---> 28 or 29 days ---> will result in 1 (logical since a month passed)
31st Jan to 1st Feb ---> 1 day ---> will result in 0 (obvious but the mentioned answer in the post results in 1 month)

Difference in Months between two dates in JavaScript:
start_date = new Date(year, month, day); //Create start date object by passing appropiate argument
end_date = new Date(new Date(year, month, day)
total months between start_date and end_date :
total_months = (end_date.getFullYear() - start_date.getFullYear())*12 + (end_date.getMonth() - start_date.getMonth())

I know this is really late, but posting it anyway just in case it helps others. Here is a function I came up with that seems to do a good job of counting differences in months between two dates. It is admittedly a great deal raunchier than Mr.Crowder's, but provides more accurate results by stepping through the date object. It is in AS3 but you should just be able to drop the strong typing and you'll have JS. Feel free to make it nicer looking anyone out there!
function countMonths ( startDate:Date, endDate:Date ):int
{
var stepDate:Date = new Date;
stepDate.time = startDate.time;
var monthCount:int;
while( stepDate.time <= endDate.time ) {
stepDate.month += 1;
monthCount += 1;
}
if ( stepDate != endDate ) {
monthCount -= 1;
}
return monthCount;
}

You could also consider this solution, this function returns the month difference in integer or number
Passing the start date as the first or last param, is fault tolerant. Meaning, the function would still return the same value.
const diffInMonths = (end, start) => {
var timeDiff = Math.abs(end.getTime() - start.getTime());
return Math.round(timeDiff / (2e3 * 3600 * 365.25));
}
const result = diffInMonths(new Date(2015, 3, 28), new Date(2010, 1, 25));
// shows month difference as integer/number
console.log(result);

To expand on #T.J.'s answer, if you're looking for simple months, rather than full calendar months, you could just check if d2's date is greater than or equal to than d1's. That is, if d2 is later in its month than d1 is in its month, then there is 1 more month. So you should be able to just do this:
function monthDiff(d1, d2) {
var months;
months = (d2.getFullYear() - d1.getFullYear()) * 12;
months -= d1.getMonth() + 1;
months += d2.getMonth();
// edit: increment months if d2 comes later in its month than d1 in its month
if (d2.getDate() >= d1.getDate())
months++
// end edit
return months <= 0 ? 0 : months;
}
monthDiff(
new Date(2008, 10, 4), // November 4th, 2008
new Date(2010, 2, 12) // March 12th, 2010
);
// Result: 16; 4 Nov – 4 Dec '08, 4 Dec '08 – 4 Dec '09, 4 Dec '09 – 4 March '10
This doesn't totally account for time issues (e.g. 3 March at 4:00pm and 3 April at 3:00pm), but it's more accurate and for just a couple lines of code.

Consider each date in terms of months, then subtract to find the difference.
var past_date = new Date('11/1/2014');
var current_date = new Date();
var difference = (current_date.getFullYear()*12 + current_date.getMonth()) - (past_date.getFullYear()*12 + past_date.getMonth());
This will get you the difference of months between the two dates, ignoring the days.

There are two approaches, mathematical & quick, but subject to vagaries in the calendar, or iterative & slow, but handles all the oddities (or at least delegates handling them to a well-tested library).
If you iterate through the calendar, incrementing the start date by one month & seeing if we pass the end date. This delegates anomaly-handling to the built-in Date() classes, but could be slow IF you're doing this for a large number of dates. James' answer takes this approach. As much as I dislike the idea, I think this is the "safest" approach, and if you're only doing one calculation, the performance difference really is negligible. We tend to try to over-optimize tasks which will only be performed once.
Now, if you're calculating this function on a dataset, you probably don't want to run that function on each row (or god forbid, multiple times per record). In that case, you can use almost any of the other answers here except the accepted answer, which is just wrong (difference between new Date() and new Date() is -1)?
Here's my stab at a mathematical-and-quick approach, which accounts for differing month lengths and leap years. You really should only use a function like this if you'll be applying this to a dataset (doing this calculation over & over). If you just need to do it once, use James' iterative approach above, as you're delegating handling all the (many) exceptions to the Date() object.
function diffInMonths(from, to){
var months = to.getMonth() - from.getMonth() + (12 * (to.getFullYear() - from.getFullYear()));
if(to.getDate() < from.getDate()){
var newFrom = new Date(to.getFullYear(),to.getMonth(),from.getDate());
if (to < newFrom && to.getMonth() == newFrom.getMonth() && to.getYear() %4 != 0){
months--;
}
}
return months;
}

Calculate the difference between two dates include fraction of month (days).
var difference = (date2.getDate() - date1.getDate()) / 30 +
date2.getMonth() - date1.getMonth() +
(12 * (date2.getFullYear() - date1.getFullYear()));
For example:
date1: 24/09/2015 (24th Sept 2015)
date2: 09/11/2015 (9th Nov 2015)
the difference: 2.5 (months)

Here you go other approach with less looping:
calculateTotalMonthsDifference = function(firstDate, secondDate) {
var fm = firstDate.getMonth();
var fy = firstDate.getFullYear();
var sm = secondDate.getMonth();
var sy = secondDate.getFullYear();
var months = Math.abs(((fy - sy) * 12) + fm - sm);
var firstBefore = firstDate > secondDate;
firstDate.setFullYear(sy);
firstDate.setMonth(sm);
firstBefore ? firstDate < secondDate ? months-- : "" : secondDate < firstDate ? months-- : "";
return months;
}

This should work fine:
function monthDiff(d1, d2) {
var months;
months = (d2.getFullYear() - d1.getFullYear()) * 12;
months += d2.getMonth() - d1.getMonth();
return months;
}

Number Of Months When Day & Time Doesn't Matter
In this case, I'm not concerned with full months, part months, how long a month is, etc. I just need to know the number of months. A relevant real world case would be where a report is due every month, and I need to know how many reports there should be.
Example:
January = 1 month
January - February = 2 months
November - January = 3 months
This is an elaborated code example to show where the numbers are going.
Let's take 2 timestamps that should result in 4 months
November 13, 2019's timestamp: 1573621200000
February 20, 2020's timestamp: 1582261140000
May be slightly different with your timezone / time pulled. The day, minutes, and seconds don't matter and can be included in the timestamp, but we will disregard it with our actual calculation.
Step 1: convert the timestamp to a JavaScript date
let dateRangeStartConverted = new Date(1573621200000);
let dateRangeEndConverted = new Date(1582261140000);
Step 2: get integer values for the months / years
let startingMonth = dateRangeStartConverted.getMonth();
let startingYear = dateRangeStartConverted.getFullYear();
let endingMonth = dateRangeEndConverted.getMonth();
let endingYear = dateRangeEndConverted.getFullYear();
This gives us
Starting month: 11
Starting Year: 2019
Ending month: 2
Ending Year: 2020
Step 3: Add (12 * (endYear - startYear)) + 1 to the ending month.
This makes our starting month stay at 11
This makes our ending month equal 15 2 + (12 * (2020 - 2019)) + 1 = 15
Step 4: Subtract the months
15 - 11 = 4; we get our 4 month result.
29 Month Example Example
November 2019 through March 2022 is 29 months. If you put these into an excel spreadsheet, you will see 29 rows.
Our starting month is 11
Our ending month is 40 3 + (12 * (2022-2019)) + 1
40 - 11 = 29

function calcualteMonthYr(){
var fromDate =new Date($('#txtDurationFrom2').val()); //date picker (text fields)
var toDate = new Date($('#txtDurationTo2').val());
var months=0;
months = (toDate.getFullYear() - fromDate.getFullYear()) * 12;
months -= fromDate.getMonth();
months += toDate.getMonth();
if (toDate.getDate() < fromDate.getDate()){
months--;
}
$('#txtTimePeriod2').val(months);
}

Following code returns full months between two dates by taking nr of days of partial months into account as well.
var monthDiff = function(d1, d2) {
if( d2 < d1 ) {
var dTmp = d2;
d2 = d1;
d1 = dTmp;
}
var months = (d2.getFullYear() - d1.getFullYear()) * 12;
months -= d1.getMonth() + 1;
months += d2.getMonth();
if( d1.getDate() <= d2.getDate() ) months += 1;
return months;
}
monthDiff(new Date(2015, 01, 20), new Date(2015, 02, 20))
> 1
monthDiff(new Date(2015, 01, 20), new Date(2015, 02, 19))
> 0
monthDiff(new Date(2015, 01, 20), new Date(2015, 01, 22))
> 0

function monthDiff(d1, d2) {
var months, d1day, d2day, d1new, d2new, diffdate,d2month,d2year,d1maxday,d2maxday;
months = (d2.getFullYear() - d1.getFullYear()) * 12;
months -= d1.getMonth() + 1;
months += d2.getMonth();
months = (months <= 0 ? 0 : months);
d1day = d1.getDate();
d2day = d2.getDate();
if(d1day > d2day)
{
d2month = d2.getMonth();
d2year = d2.getFullYear();
d1new = new Date(d2year, d2month-1, d1day,0,0,0,0);
var timeDiff = Math.abs(d2.getTime() - d1new.getTime());
diffdate = Math.abs(Math.ceil(timeDiff / (1000 * 3600 * 24)));
d1new = new Date(d2year, d2month, 1,0,0,0,0);
d1new.setDate(d1new.getDate()-1);
d1maxday = d1new.getDate();
months += diffdate / d1maxday;
}
else
{
if(!(d1.getMonth() == d2.getMonth() && d1.getFullYear() == d2.getFullYear()))
{
months += 1;
}
diffdate = d2day - d1day + 1;
d2month = d2.getMonth();
d2year = d2.getFullYear();
d2new = new Date(d2year, d2month + 1, 1, 0, 0, 0, 0);
d2new.setDate(d2new.getDate()-1);
d2maxday = d2new.getDate();
months += diffdate / d2maxday;
}
return months;
}

below logic will fetch difference in months
(endDate.getFullYear()*12+endDate.getMonth())-(startDate.getFullYear()*12+startDate.getMonth())

function monthDiff(date1, date2, countDays) {
countDays = (typeof countDays !== 'undefined') ? countDays : false;
if (!date1 || !date2) {
return 0;
}
let bigDate = date1;
let smallDate = date2;
if (date1 < date2) {
bigDate = date2;
smallDate = date1;
}
let monthsCount = (bigDate.getFullYear() - smallDate.getFullYear()) * 12 + (bigDate.getMonth() - smallDate.getMonth());
if (countDays && bigDate.getDate() < smallDate.getDate()) {
--monthsCount;
}
return monthsCount;
}

This is the simplest solution I could find. This will directly return the number of months. Although, it always gives an absolute value.
new Date(new Date(d2) - new Date(d1)).getMonth();
For non-absolute values, you can use the following solution:
function diff_months(startDate, endDate) {
let diff = new Date( new Date(endDate) - new Date(startDate) ).getMonth();
return endDate >= startDate ? diff : -diff;
}

See what I use:
function monthDiff() {
var startdate = Date.parseExact($("#startingDate").val(), "dd/MM/yyyy");
var enddate = Date.parseExact($("#endingDate").val(), "dd/MM/yyyy");
var months = 0;
while (startdate < enddate) {
if (startdate.getMonth() === 1 && startdate.getDate() === 28) {
months++;
startdate.addMonths(1);
startdate.addDays(2);
} else {
months++;
startdate.addMonths(1);
}
}
return months;
}

It also counts the days and convert them in months.
function monthDiff(d1, d2) {
var months;
months = (d2.getFullYear() - d1.getFullYear()) * 12; //calculates months between two years
months -= d1.getMonth() + 1;
months += d2.getMonth(); //calculates number of complete months between two months
day1 = 30-d1.getDate();
day2 = day1 + d2.getDate();
months += parseInt(day2/30); //calculates no of complete months lie between two dates
return months <= 0 ? 0 : months;
}
monthDiff(
new Date(2017, 8, 8), // Aug 8th, 2017 (d1)
new Date(2017, 12, 12) // Dec 12th, 2017 (d2)
);
//return value will be 4 months

getMonthDiff(d1, d2) {
var year1 = dt1.getFullYear();
var year2 = dt2.getFullYear();
var month1 = dt1.getMonth();
var month2 = dt2.getMonth();
var day1 = dt1.getDate();
var day2 = dt2.getDate();
var months = month2 - month1;
var years = year2 -year1
days = day2 - day1;
if (days < 0) {
months -= 1;
}
if (months < 0) {
months += 12;
}
return months + years*!2;
}

Any value is returned along with its absolute value.
function differenceInMonths(firstDate, secondDate) {
if (firstDate > secondDate) [firstDate, secondDate] = [secondDate, firstDate];
let diffMonths = (secondDate.getFullYear() - firstDate.getFullYear()) * 12;
diffMonths -= firstDate.getMonth();
diffMonths += secondDate.getMonth();
return diffMonths;
}

The following code snippet helped me to find months between two dates
Find Months Count Between two dates JS
Months Between two dates JS
Code Snippet
function diff_months_count(startDate, endDate) {
var months;
var d1 = new Date(startDate);
var d2 = new Date(endDate);
months = (d2.getFullYear() - d1.getFullYear()) * 12;
months -= d1.getMonth();
months += d2.getMonth();
return months <= 0 ? 0 : months;
}

#Here is a nice piece of code i wrote for getting number of days and months
from given dates
[1]: jsfiddle link
/**
* Date a end day
* Date b start day
* #param DateA Date #param DateB Date
* #returns Date difference
*/
function getDateDifference(dateA, DateB, type = 'month') {
const END_DAY = new Date(dateA)
const START_DAY = new Date(DateB)
let calculatedDateBy
let returnDateDiff
if (type === 'month') {
const startMonth = START_DAY.getMonth()
const endMonth = END_DAY.getMonth()
calculatedDateBy = startMonth - endMonth
returnDateDiff = Math.abs(
calculatedDateBy + 12 * (START_DAY.getFullYear() - END_DAY.getFullYear())
)
} else {
calculatedDateBy = Math.abs(START_DAY - END_DAY)
returnDateDiff = Math.ceil(calculatedDateBy / (1000 * 60 * 60 * 24))
}
const out = document.getElementById('output')
out.innerText = returnDateDiff
return returnDateDiff
}
// Gets number of days from given dates
/* getDateDifference('2022-03-31','2022-04-08','day') */
// Get number of months from given dates
getDateDifference('2021-12-02','2022-04-08','month')
<div id="output"> </div>

anyVar = (((DisplayTo.getFullYear() * 12) + DisplayTo.getMonth()) - ((DisplayFrom.getFullYear() * 12) + DisplayFrom.getMonth()));

One approach would be to write a simple Java Web Service (REST/JSON) that uses JODA library
http://joda-time.sourceforge.net/faq.html#datediff
to calculate difference between two dates and call that service from javascript.
This assumes your back end is in Java.