Develop Reference

Javascript: convert number of days to years, months and days

I have days, months and years. I'm doing calculations between them. That means I have to divide 2 years 3 months and 10 days by 1/4. Now i have following code:
const getCurrentDate = moment().format("YYYY-MM-DD");
const timeEnd = moment(moment(DefEndDate).format("YYYY-MM-DD"));
const diff = timeEnd.diff(getCurrentDate);
const diffDuration = moment.duration(diff);
const diffCount = moment.duration(diff).asDays();
console.log(diffCount);
console.log("Years:", diffDuration.years());
console.log("Month:", diffDuration.months());
console.log("Days:", diffDuration.days());
const diffCount = moment.duration(diff).asDays(); //Get it as days
const [unserve, setUnserve] = useState(''); //set value to variable
const res = unserve.split('/'); //split 1/4 to 1.4
const x = parseFloat(res[0] + "." + res[1]); //convert it to float
var quotient = Math.floor(diffCount/x); //calculate
console.log(quotient);
//returned 832 / 1.4 = 594 days
Now I need to return the output number (days) to the year, month and day. I can't do that. How do I convert? And another question is, can this way be the optimal solution?

I can't decide whether what you really want to do is divide a date range in to a fixed number of periods with equal days, or to start with a date, add a period in years, months and days to get an end date, then divide that into equal periods.
The following assumes the latter.
I have to divide 2 years 3 months and 10 days by 1/4
The number of days covered by that period varies depending the dates it is to and from, so you have to start with the start and end dates of the range.
In your code:
const getCurrentDate = moment().format("YYYY-MM-DD");
Sets getCurrentDate to a string like 2020-02-11.
const timeEnd = moment(moment(DefEndDate).format("YYYY-MM-DD"));
Creates a moment object from the string value of getCurrentDate and sets timeEnd to another string.
const diff = timeEnd.diff(getCurrentDate);
This attempts to call the diff method of timeEnd, which is a string. Strings don't have a diff method so the expression returns undefined, attempting to call it throws an error something like TypeError: '2020-02-11'.diff is not a function.
The rest of your code seems to be based on a algorithm
If you have a predetermined period in years, months days, etc. you can start with a start date, add the period, then get the number of days difference. Divide that difference by the number of periods you want, then add that sequentially to get the various end dates.
The following example uses moment.js since that's what you appear to be using, however a version without a library is about the same difficulty. It returns an array of dates, starting with the start date so there is one more date than periods.
function getDates(
start = new Date(),
years = 0,
months = 0,
days = 0,
parts = 1) {
// Get start and end as moment objects
let m = moment(start).startOf('day');
let end = moment(m);
end.add({years:years, months:months, days:days});
// Get days difference and number of days to add for each period
let daysDiff = end.diff(m, 'days');
let f = daysDiff / parts;
let dayArray = [m.format('YYYY-MM-DD')];
let i = 0;
while ((f * ++i) <= daysDiff) {
let d = moment(m).add(f * i, 'days')
dayArray.push(d.format('YYYY-MM-DD'));
}
return dayArray;
}
// Get dates for 4 even periods over 2 years, 3 months and
// 10 days from today
console.log(getDates(new Date(), 2, 3, 10, 4));
<script src="https://cdnjs.cloudflare.com/ajax/libs/moment.js/2.24.0/moment.min.js"></script>
As the sub–period length is nearly always not an even number of days (in the above case it's about 207.5), I've allowed formatting to effectively truncate the decimal part of a day. You might want to use some other rounding algorithm that more evenly distributes the remainder.
If, on the other hand, you have start and end dates and want an equal number of periods, the following is much simpler (and doesn't use a library):
// Helpers - use UTC do avoid DST issues
function toUTCDate(s) {
let b = s.split(/\D/);
return new Date(Date.UTC(b[0], --b[1], b[2]));
}
function formatUTC(date) {
return date.toISOString().substr(0, 10);
}
/* #param {string} start - date string in format 'YYYY-MM-DD'
** #param {string} end - date string in format 'YYYY-MM-DD'
** #param (numbe} n - number of required periods
** #returns {Array} array of date strings in format 'YYYY-MM-DD'
*/
function periods(start, end, n) {
let s = toUTCDate(start);
let e = toUTCDate(end);
let diff = e - s;
let f = diff / n;
let result = [formatUTC(s)];
// Allow for rounding of decimal f in comparison
while (e - s > n) {
s.setTime(s.getTime() + f);
result.push(formatUTC(s))
}
return result;
}
console.log(periods('2020-02-09','2022-05-19', 4));
The two methods produce slightly different results, you'll need to work out if that matters or not.

Long time ago I used such code to convert from seconds to hours, minutes and seconds:
var hours = Math.floor(sec / 3600);
var minutes = Math.floor((sec - hours * 3600) / 60);
var seconds = Math.round(sec - hours * 3600 - minutes * 60);
Math with days, months and years should be somewhat similar but my rusty brain doesn't want to think about it
UPD
This function seems to do what you need
const daysFmt = days => {
const years = Math.floor(days / 365);
const months = Math.floor((days - years * 365) / 30);
const d = Math.round(days - years * 365 - months * 30);
let res = [];
if (years > 0) {
res.push(years + ' y');
}
if (months > 0) {
res.push(months + ' m');
}
if (d > 0) {
res.push(d + ' d');
}
return res.join(', ');
}
But this solution has one nuance: it assumes that month = 30 days. You might want to add if statement to return '1 m' if input is 31 or just return number of days if it is less than 32. Test results:
daysFmt(31);
"1 m, 1 d"
daysFmt(180);
"6 m"
daysFmt(185);
"6 m, 5 d"
daysFmt(356);
"11 m, 26 d"
daysFmt(365);
"1 y"
daysFmt(420);
"1 y, 1 m, 25 d"
daysFmt(3650);
"10 y"
daysFmt(3685);
"10 y, 1 m, 5 d"

How do I get the number of days for a duration where the number of hours exceeds 24 in google sheets?

I am using google sheets where there is a duration value of 69:41:00 where it's 69 hours, 41 minutes, 0 secs. There doesn't seem to be a function to convert this to days, hours and minutes so I did some searching and some had suggested a custom function. Not sure exactly how it works but made some changes from the original to fit what I needed. The code below:
/**
* Format Duration to Days,Hours,Minutes
*
* #param {duration} input value.
* #return Days,Hours,Minutes.
* #customfunction
*/
function FormatDuration(duration) {
// Retrieve the hours and minutes
var hrs = duration.getHours();
var days = Math.floor(hrs/24);
var hours = hrs % 24;
var mins = duration.getMinutes();
// Convert the result to a number to use in calculations
var result = days + 'd ' + hours + ' h '+ mins+' min';
return result;
}
The result should be 2d 21h 44 min but instead I got 0d 21 h 35 min. Am I doing something wrong here?

I was going to add, why don't you just use a custom format of
ʺd\d hh\h mm\mʺ ?
This works fine in Excel but not in GS because it uses a different base for dates so duration like 69:41:00 would be interpreted as 1/1/1900 21:41 and the days are not correct. So you would have to break it down into days (whole numbers) and hours+minutes (fractions of a day) like this
=text(int(A1),ʺ#0\d ʺ)&text(mod(A1,1),ʺHH\h MM\mʺ)
You can make it work in Google Scripts if you want to by adjusting the date - should work OK for durations up to 1 month.
The reason for adding 2 to the date is that a time like 03:21:00 (less than a day) is seen as a date - namely 30th December 1899 ! So I add 2 to it to make it 1st January 1900. However, now the day part of the date is 1 and I want it to be zero. So I have to subtract 1 from the day further down.
This strange behaviour is probably why you're advised to do it the other way and work in milliseconds, but I was just interested to see if there was a way of making the original code work.
/**
* Format Duration to Days,Hours,Minutes
*
* #param {duration} input value.
* #return Days,Hours,Minutes.
* #customfunction
*/
function FormatDuration(duration) {
// Add 2 days to the date
var date=new Date(duration.setDate(duration.getDate()+2));
Logger.log(date.getDate());
var hours = duration.getHours();
// Take 1 off the day part of the date
var days = date.getDate()-1;
var mins = duration.getMinutes();
// Convert the result to a number to use in calculations
var result = days + 'd ' + hours + ' h '+ mins+' min';
return result;
}

function(durations){
var timeArr = durations.split(':'); //["69","41","00"]
//your code
}
getHours is a method of object Date.
var t = new Date;
t.getHours();

How do you expect to get more than 24hours from a Date object? It is not the same as what you expect as Duration. Date is for points of time in calendar, so at most you'd get the 23:59:59 of any day. You can get date2 - date1 = milliseconds diff, and work on it, as following;
function FormatDuration(date1, date2) {
var milliseconds = date2 - date1;
var mins = Math.floor((milliseconds / (1000*60)) % 60);
var hours = Math.floor((milliseconds / (1000*60*60)) % 24);
var days = Math.floor(milliseconds / (1000*60*60*24));
var result = days + ' d ' + hours + ' h '+ mins + ' min';
console.log(result);
}
FormatDuration(new Date(2000, 5, 1, 5, 13, 0, 0),
new Date(2000, 5, 2, 15, 31, 0, 0))
You can find more details here

Show week number with Javascript?

I have the following code that is used to show the name of the current day, followed by a set phrase.
<script type="text/javascript">
<!--
// Array of day names
var dayNames = new Array(
"It's Sunday, the weekend is nearly over",
"Yay! Another Monday",
"Hello Tuesday, at least you're not Monday",
"It's Wednesday. Halfway through the week already",
"It's Thursday.",
"It's Friday - Hurray for the weekend",
"Saturday Night Fever");
var now = new Date();
document.write(dayNames[now.getDay()] + ".");
// -->
</script>
What I would like to do is have the current week number in brackets after the phrase. I have found the following code:
Date.prototype.getWeek = function() {
var onejan = new Date(this.getFullYear(),0,1);
return Math.ceil((((this - onejan) / 86400000) + onejan.getDay()+1)/7);
}
Which was taken from http://javascript.about.com/library/blweekyear.htm but I have no idea how to add it to existing javascript code.

Simply add it to your current code, then call (new Date()).getWeek()
<script>
Date.prototype.getWeek = function() {
var onejan = new Date(this.getFullYear(), 0, 1);
return Math.ceil((((this - onejan) / 86400000) + onejan.getDay() + 1) / 7);
}
var weekNumber = (new Date()).getWeek();
var dayNames = ['Sunday', 'Monday', 'Tuesday', 'Wednesday', 'Thursday', 'Friday', 'Saturday'];
var now = new Date();
document.write(dayNames[now.getDay()] + " (" + weekNumber + ").");
</script>

In case you already use jQuery-UI (specifically datepicker):
Date.prototype.getWeek = function () { return $.datepicker.iso8601Week(this); }
Usage:
var myDate = new Date();
myDate.getWeek();
More here: UI/Datepicker/iso8601Week
I realize this isn't a general solution as it incurs a dependency. However, considering the popularity of jQuery-UI this might just be a simple fit for someone - as it was for me.
If you don't use jQuery-UI and have no intention of adding the dependency. You could just copy their iso8601Week() implementation since it is written in pure JavaScript without complex dependencies:
// Determine the week of the year (local timezone) based on the ISO 8601 definition.
Date.prototype.iso8601Week = function () {
// Create a copy of the current date, we don't want to mutate the original
const date = new Date(this.getTime());
// Find Thursday of this week starting on Monday
date.setDate(date.getDate() + 4 - (date.getDay() || 7));
const thursday = date.getTime();
// Find January 1st
date.setMonth(0); // January
date.setDate(1); // 1st
const jan1st = date.getTime();
// Round the amount of days to compensate for daylight saving time
const days = Math.round((thursday - jan1st) / 86400000); // 1 day = 86400000 ms
return Math.floor(days / 7) + 1;
};
console.log(new Date().iso8601Week());
console.log(new Date("2020-01-01T00:00").iso8601Week());
console.log(new Date("2021-01-01T00:00").iso8601Week());
console.log(new Date("2022-01-01T00:00").iso8601Week());
console.log(new Date("2023-12-31T00:00").iso8601Week());
console.log(new Date("2024-12-31T00:00").iso8601Week());

Consider using my implementation of "Date.prototype.getWeek", think is more accurate than the others i have seen here :)
Date.prototype.getWeek = function(){
// We have to compare against the first monday of the year not the 01/01
// 60*60*24*1000 = 86400000
// 'onejan_next_monday_time' reffers to the miliseconds of the next monday after 01/01
var day_miliseconds = 86400000,
onejan = new Date(this.getFullYear(),0,1,0,0,0),
onejan_day = (onejan.getDay()==0) ? 7 : onejan.getDay(),
days_for_next_monday = (8-onejan_day),
onejan_next_monday_time = onejan.getTime() + (days_for_next_monday * day_miliseconds),
// If one jan is not a monday, get the first monday of the year
first_monday_year_time = (onejan_day>1) ? onejan_next_monday_time : onejan.getTime(),
this_date = new Date(this.getFullYear(), this.getMonth(),this.getDate(),0,0,0),// This at 00:00:00
this_time = this_date.getTime(),
days_from_first_monday = Math.round(((this_time - first_monday_year_time) / day_miliseconds));
var first_monday_year = new Date(first_monday_year_time);
// We add 1 to "days_from_first_monday" because if "days_from_first_monday" is *7,
// then 7/7 = 1, and as we are 7 days from first monday,
// we should be in week number 2 instead of week number 1 (7/7=1)
// We consider week number as 52 when "days_from_first_monday" is lower than 0,
// that means the actual week started before the first monday so that means we are on the firsts
// days of the year (ex: we are on Friday 01/01, then "days_from_first_monday"=-3,
// so friday 01/01 is part of week number 52 from past year)
// "days_from_first_monday<=364" because (364+1)/7 == 52, if we are on day 365, then (365+1)/7 >= 52 (Math.ceil(366/7)=53) and thats wrong
return (days_from_first_monday>=0 && days_from_first_monday<364) ? Math.ceil((days_from_first_monday+1)/7) : 52;
}
You can check my public repo here https://bitbucket.org/agustinhaller/date.getweek (Tests included)

If you want something that works and is future-proof, use a library like MomentJS.
moment(date).week();
moment(date).isoWeek()
http://momentjs.com/docs/#/get-set/week/

It looks like this function I found at weeknumber.net is pretty accurate and easy to use.
// This script is released to the public domain and may be used, modified and
// distributed without restrictions. Attribution not necessary but appreciated.
// Source: http://weeknumber.net/how-to/javascript
// Returns the ISO week of the date.
Date.prototype.getWeek = function() {
var date = new Date(this.getTime());
date.setHours(0, 0, 0, 0);
// Thursday in current week decides the year.
date.setDate(date.getDate() + 3 - (date.getDay() + 6) % 7);
// January 4 is always in week 1.
var week1 = new Date(date.getFullYear(), 0, 4);
// Adjust to Thursday in week 1 and count number of weeks from date to week1.
return 1 + Math.round(((date.getTime() - week1.getTime()) / 86400000 - 3 + (week1.getDay() + 6) % 7) / 7);
}
If you're lucky like me and need to find the week number of the month a little adjust will do it:
// Returns the week in the month of the date.
Date.prototype.getWeekOfMonth = function() {
var date = new Date(this.getTime());
date.setHours(0, 0, 0, 0);
// Thursday in current week decides the year.
date.setDate(date.getDate() + 3 - (date.getDay() + 6) % 7);
// January 4 is always in week 1.
var week1 = new Date(date.getFullYear(), date.getMonth(), 4);
// Adjust to Thursday in week 1 and count number of weeks from date to week1.
return 1 + Math.round(((date.getTime() - week1.getTime()) / 86400000 - 3 + (week1.getDay() + 6) % 7) / 7);
}

If you already use Angular, then you could profit $filter('date').
For example:
var myDate = new Date();
var myWeek = $filter('date')(myDate, 'ww');

By adding the snippet you extend the Date object.
Date.prototype.getWeek = function() {
var onejan = new Date(this.getFullYear(),0,1);
return Math.ceil((((this - onejan) / 86400000) + onejan.getDay()+1)/7);
}
If you want to use this in multiple pages you can add this to a seperate js file which must be loaded first before your other scripts executes. With other scripts I mean the scripts which uses the getWeek() method.

All the proposed approaches may give wrong results because they don’t take into account summer/winter time changes. Rather than calculating the number of days between two dates using the constant of 86’400’000 milliseconds, it is better to use an approach like the following one:
getDaysDiff = function (dateObject0, dateObject1) {
if (dateObject0 >= dateObject1) return 0;
var d = new Date(dateObject0.getTime());
var nd = 0;
while (d <= dateObject1) {
d.setDate(d.getDate() + 1);
nd++;
}
return nd-1;
};

I was coding in the dark (a challenge) and couldn't lookup, bring in any dependencies or test my code.
I forgot what round up was called (Math.celi) So I wanted to be extra sure i got it right and came up with this code instead.
var elm = document.createElement('input')
elm.type = 'week'
elm.valueAsDate = new Date()
var week = elm.value.split('W').pop()
console.log(week)
Just a proof of concept of how you can get the week in any other way
But still i recommend any other solution that isn't required by the DOM.

With that code you can simply;
document.write(dayNames[now.getDay()] + " (" + now.getWeek() + ").");
(You will need to paste the getWeek function above your current script)

You could find this fiddle useful. Just finished.
https://jsfiddle.net/dnviti/ogpt920w/
Code below also:
/**
* Get the ISO week date week number
*/
Date.prototype.getWeek = function () {
// Create a copy of this date object
var target = new Date(this.valueOf());
// ISO week date weeks start on monday
// so correct the day number
var dayNr = (this.getDay() + 6) % 7;
// ISO 8601 states that week 1 is the week
// with the first thursday of that year.
// Set the target date to the thursday in the target week
target.setDate(target.getDate() - dayNr + 3);
// Store the millisecond value of the target date
var firstThursday = target.valueOf();
// Set the target to the first thursday of the year
// First set the target to january first
target.setMonth(0, 1);
// Not a thursday? Correct the date to the next thursday
if (target.getDay() != 4) {
target.setMonth(0, 1 + ((4 - target.getDay()) + 7) % 7);
}
// The weeknumber is the number of weeks between the
// first thursday of the year and the thursday in the target week
return 1 + Math.ceil((firstThursday - target) / 604800000); // 604800000 = 7 * 24 * 3600 * 1000
}
/**
* Get the ISO week date year number
*/
Date.prototype.getWeekYear = function ()
{
// Create a new date object for the thursday of this week
var target = new Date(this.valueOf());
target.setDate(target.getDate() - ((this.getDay() + 6) % 7) + 3);
return target.getFullYear();
}
/**
* Convert ISO week number and year into date (first day of week)
*/
var getDateFromISOWeek = function(w, y) {
var simple = new Date(y, 0, 1 + (w - 1) * 7);
var dow = simple.getDay();
var ISOweekStart = simple;
if (dow <= 4)
ISOweekStart.setDate(simple.getDate() - simple.getDay() + 1);
else
ISOweekStart.setDate(simple.getDate() + 8 - simple.getDay());
return ISOweekStart;
}
var printDate = function(){
/*var dateString = document.getElementById("date").value;
var dateArray = dateString.split("/");*/ // use this if you have year-week in the same field
var dateInput = document.getElementById("date").value;
if (dateInput == ""){
var date = new Date(); // get today date object
}
else{
var date = new Date(dateInput); // get date from field
}
var day = ("0" + date.getDate()).slice(-2); // get today day
var month = ("0" + (date.getMonth() + 1)).slice(-2); // get today month
var fullDate = date.getFullYear()+"-"+(month)+"-"+(day) ; // get full date
var year = date.getFullYear();
var week = ("0" + (date.getWeek())).slice(-2);
var locale= "it-it";
document.getElementById("date").value = fullDate; // set input field
document.getElementById("year").value = year;
document.getElementById("week").value = week; // this prototype has been written above
var fromISODate = getDateFromISOWeek(week, year);
var fromISODay = ("0" + fromISODate.getDate()).slice(-2);
var fromISOMonth = ("0" + (fromISODate.getMonth() + 1)).slice(-2);
var fromISOYear = date.getFullYear();
// Use long to return month like "December" or short for "Dec"
//var monthComplete = fullDate.toLocaleString(locale, { month: "long" });
var formattedDate = fromISODay + "-" + fromISOMonth + "-" + fromISOYear;
var element = document.getElementById("fullDate");
element.value = formattedDate;
}
printDate();
document.getElementById("convertToDate").addEventListener("click", printDate);
*{
font-family: consolas
}
<label for="date">Date</label>
<input type="date" name="date" id="date" style="width:130px;text-align:center" value="" />
<br /><br />
<label for="year">Year</label>
<input type="year" name="year" id="year" style="width:40px;text-align:center" value="" />
-
<label for="week">Week</label>
<input type="text" id="week" style="width:25px;text-align:center" value="" />
<br /><br />
<label for="fullDate">Full Date</label>
<input type="text" id="fullDate" name="fullDate" style="width:80px;text-align:center" value="" />
<br /><br />
<button id="convertToDate">
Convert Date
</button>
It's pure JS.
There are a bunch of date functions inside that allow you to convert date into week number and viceversa :)

Luxon is an other alternative. Luxon date objects have a weekNumber property:
let week = luxon.DateTime.fromString("2022-04-01", "yyyy-MM-dd").weekNumber;
console.log(week);
<script src="https://cdnjs.cloudflare.com/ajax/libs/luxon/3.0.1/luxon.min.js"></script>

I've tried using code from all of the answers above, and all return week #52 for the first of January. So I decided to write my own, which calculates the week number correctly.
Week numeration starts from 0
Maybe it's a bad taste to use a loop, or the result can be cached somewhere to prevent repeating the same calculations if the function is called often enough. Well, I have made this for myself, and it does what I need it to do.
Date.prototype.getWeek = function() {
// debugger
let msWeek = 604800000; // Week in milliseconds
let msDay = 86400000; // Day in milliseconds
let year = this.getFullYear(); // Get the year
//let month = this.getMonth(); // Month
let oneDate = new Date(year, 0, 1); // Create a new date based on THIS year
let temp = oneDate.getDay(); // Ordinal of the first day
let getFirstDay = (temp === 0) ? 6 : temp - 1; // Ordinal of the first day of the current month (0-MO, 6-SU)
let countWeek = 0;
// Test to confirm week
oneDate = new Date(oneDate.getTime() + msDay*(7 - getFirstDay));
if(oneDate.getTime() > this.getTime()){
return countWeek;
}
// Increment loop
while(true){
oneDate = new Date(oneDate.getTime() + msWeek); // Add a week and check
if(oneDate.getTime() > this.getTime()) break;
countWeek++;
}
return countWeek + 1;
}
let s1 = new Date('2022-01-01'); console.log(s1.getWeek());
let s2 = new Date('2023-01-01'); console.log(s2.getWeek());
let s22 = new Date('2023-01-02'); console.log(s22.getWeek());
let s3 = new Date('2024-01-01'); console.log(s3.getWeek());
let s4 = new Date('2025-01-01'); console.log(s4.getWeek());
let s5 = new Date('2022-02-28'); console.log(s5.getWeek());
let s6 = new Date('2022-12-31'); console.log(s6.getWeek());
let s7 = new Date('2024-12-31'); console.log(s7.getWeek());

Some of the code I see in here fails with years like 2016, in which week 53 jumps to week 2.
Here is a revised and working version:
Date.prototype.getWeek = function() {
// Create a copy of this date object
var target = new Date(this.valueOf());
// ISO week date weeks start on monday, so correct the day number
var dayNr = (this.getDay() + 6) % 7;
// Set the target to the thursday of this week so the
// target date is in the right year
target.setDate(target.getDate() - dayNr + 3);
// ISO 8601 states that week 1 is the week with january 4th in it
var jan4 = new Date(target.getFullYear(), 0, 4);
// Number of days between target date and january 4th
var dayDiff = (target - jan4) / 86400000;
if(new Date(target.getFullYear(), 0, 1).getDay() < 5) {
// Calculate week number: Week 1 (january 4th) plus the
// number of weeks between target date and january 4th
return 1 + Math.ceil(dayDiff / 7);
}
else { // jan 4th is on the next week (so next week is week 1)
return Math.ceil(dayDiff / 7);
}
};

Martin Schillinger's version seems to be the strictly correct one.
Since I knew I only needed it to work correctly on business week days, I went with this simpler form, based on something I found online, don't remember where:
ISOWeekday = (0 == InputDate.getDay()) ? 7 : InputDate.getDay();
ISOCalendarWeek = Math.floor( ( ((InputDate.getTime() - (new Date(InputDate.getFullYear(),0,1)).getTime()) / 86400000) - ISOWeekday + 10) / 7 );
It fails in early January on days that belong to the previous year's last week (it produces CW = 0 in those cases) but is correct for everything else.

JavaScript - get the first day of the week from current date

I need the fastest way to get the first day of the week. For example: today is the 11th of November, and a Thursday; and I want the first day of this week, which is the 8th of November, and a Monday. I need the fastest method for MongoDB map function, any ideas?

Using the getDay method of Date objects, you can know the number of day of the week (being 0=Sunday, 1=Monday, etc).
You can then subtract that number of days plus one, for example:
function getMonday(d) {
d = new Date(d);
var day = d.getDay(),
diff = d.getDate() - day + (day == 0 ? -6:1); // adjust when day is sunday
return new Date(d.setDate(diff));
}
getMonday(new Date()); // Mon Nov 08 2010

Not sure how it compares for performance, but this works.
var today = new Date();
var day = today.getDay() || 7; // Get current day number, converting Sun. to 7
if( day !== 1 ) // Only manipulate the date if it isn't Mon.
today.setHours(-24 * (day - 1)); // Set the hours to day number minus 1
// multiplied by negative 24
alert(today); // will be Monday
Or as a function:
# modifies _date_
function setToMonday( date ) {
var day = date.getDay() || 7;
if( day !== 1 )
date.setHours(-24 * (day - 1));
return date;
}
setToMonday(new Date());

CMS's answer is correct but assumes that Monday is the first day of the week.
Chandler Zwolle's answer is correct but fiddles with the Date prototype.
Other answers that add/subtract hours/minutes/seconds/milliseconds are wrong because not all days have 24 hours.
The function below is correct and takes a date as first parameter and the desired first day of the week as second parameter (0 for Sunday, 1 for Monday, etc.). Note: the hour, minutes and seconds are set to 0 to have the beginning of the day.
function firstDayOfWeek(dateObject, firstDayOfWeekIndex) {
const dayOfWeek = dateObject.getDay(),
firstDayOfWeek = new Date(dateObject),
diff = dayOfWeek >= firstDayOfWeekIndex ?
dayOfWeek - firstDayOfWeekIndex :
6 - dayOfWeek
firstDayOfWeek.setDate(dateObject.getDate() - diff)
firstDayOfWeek.setHours(0,0,0,0)
return firstDayOfWeek
}
// August 18th was a Saturday
let lastMonday = firstDayOfWeek(new Date('August 18, 2018 03:24:00'), 1)
// outputs something like "Mon Aug 13 2018 00:00:00 GMT+0200"
// (may vary according to your time zone)
document.write(lastMonday)

First / Last Day of The Week
To get the upcoming first day of the week, you can use something like so:
function getUpcomingSunday() {
const date = new Date();
const today = date.getDate();
const currentDay = date.getDay();
const newDate = date.setDate(today - currentDay + 7);
return new Date(newDate);
}
console.log(getUpcomingSunday());
Or to get the latest first day:
function getLastSunday() {
const date = new Date();
const today = date.getDate();
const currentDay = date.getDay();
const newDate = date.setDate(today - (currentDay || 7));
return new Date(newDate);
}
console.log(getLastSunday());
* Depending on your time zone, the beginning of the week doesn't has to start on Sunday, it can start on Friday, Saturday, Monday or any other day your machine is set to. Those methods will account for that.
* You can also format it using toISOString method like so: getLastSunday().toISOString()

Check out Date.js
Date.today().previous().monday()

var dt = new Date(); // current date of week
var currentWeekDay = dt.getDay();
var lessDays = currentWeekDay == 0 ? 6 : currentWeekDay - 1;
var wkStart = new Date(new Date(dt).setDate(dt.getDate() - lessDays));
var wkEnd = new Date(new Date(wkStart).setDate(wkStart.getDate() + 6));
This will work well.

I'm using this
function get_next_week_start() {
var now = new Date();
var next_week_start = new Date(now.getFullYear(), now.getMonth(), now.getDate()+(8 - now.getDay()));
return next_week_start;
}

Returns Monday 00am to Monday 00am.
const now = new Date()
const startOfWeek = new Date(now.getFullYear(), now.getMonth(), now.getDate() - now.getDay() + 1)
const endOfWeek = new Date(now.getFullYear(), now.getMonth(), startOfWeek.getDate() + 7)

This function uses the current millisecond time to subtract the current week, and then subtracts one more week if the current date is on a monday (javascript counts from sunday).
function getMonday(fromDate) {
// length of one day i milliseconds
var dayLength = 24 * 60 * 60 * 1000;
// Get the current date (without time)
var currentDate = new Date(fromDate.getFullYear(), fromDate.getMonth(), fromDate.getDate());
// Get the current date's millisecond for this week
var currentWeekDayMillisecond = ((currentDate.getDay()) * dayLength);
// subtract the current date with the current date's millisecond for this week
var monday = new Date(currentDate.getTime() - currentWeekDayMillisecond + dayLength);
if (monday > currentDate) {
// It is sunday, so we need to go back further
monday = new Date(monday.getTime() - (dayLength * 7));
}
return monday;
}
I have tested it when week spans over from one month to another (and also years), and it seems to work properly.

Good evening,
I prefer to just have a simple extension method:
Date.prototype.startOfWeek = function (pStartOfWeek) {
var mDifference = this.getDay() - pStartOfWeek;
if (mDifference < 0) {
mDifference += 7;
}
return new Date(this.addDays(mDifference * -1));
}
You'll notice this actually utilizes another extension method that I use:
Date.prototype.addDays = function (pDays) {
var mDate = new Date(this.valueOf());
mDate.setDate(mDate.getDate() + pDays);
return mDate;
};
Now, if your weeks start on Sunday, pass in a "0" for the pStartOfWeek parameter, like so:
var mThisSunday = new Date().startOfWeek(0);
Similarly, if your weeks start on Monday, pass in a "1" for the pStartOfWeek parameter:
var mThisMonday = new Date().startOfWeek(1);
Regards,

a more generalized version of this... this will give you any day in the current week based on what day you specify.
//returns the relative day in the week 0 = Sunday, 1 = Monday ... 6 = Saturday
function getRelativeDayInWeek(d,dy) {
d = new Date(d);
var day = d.getDay(),
diff = d.getDate() - day + (day == 0 ? -6:dy); // adjust when day is sunday
return new Date(d.setDate(diff));
}
var monday = getRelativeDayInWeek(new Date(),1);
var friday = getRelativeDayInWeek(new Date(),5);
console.log(monday);
console.log(friday);

Simple solution for getting the first day of the week.
With this solution, it is possible to set an arbitrary start of week (e.g. Sunday = 0, Monday = 1, Tuesday = 2, etc.).
function getBeginOfWeek(date = new Date(), startOfWeek = 1) {
const result = new Date(date);
while (result.getDay() !== startOfWeek) {
result.setDate(result.getDate() - 1);
}
return result;
}
The solution correctly wraps on months (due to Date.setDate() being used)
For startOfWeek, the same constant numbers as in Date.getDay() can be used

setDate() has issues with month boundaries that are noted in comments above. A clean workaround is to find the date difference using epoch timestamps rather than the (surprisingly counterintuitive) methods on the Date object. I.e.
function getPreviousMonday(fromDate) {
var dayMillisecs = 24 * 60 * 60 * 1000;
// Get Date object truncated to date.
var d = new Date(new Date(fromDate || Date()).toISOString().slice(0, 10));
// If today is Sunday (day 0) subtract an extra 7 days.
var dayDiff = d.getDay() === 0 ? 7 : 0;
// Get date diff in millisecs to avoid setDate() bugs with month boundaries.
var mondayMillisecs = d.getTime() - (d.getDay() + dayDiff) * dayMillisecs;
// Return date as YYYY-MM-DD string.
return new Date(mondayMillisecs).toISOString().slice(0, 10);
}

Here is my solution:
function getWeekDates(){
var day_milliseconds = 24*60*60*1000;
var dates = [];
var current_date = new Date();
var monday = new Date(current_date.getTime()-(current_date.getDay()-1)*day_milliseconds);
var sunday = new Date(monday.getTime()+6*day_milliseconds);
dates.push(monday);
for(var i = 1; i < 6; i++){
dates.push(new Date(monday.getTime()+i*day_milliseconds));
}
dates.push(sunday);
return dates;
}
Now you can pick date by returned array index.

An example of the mathematically only calculation, without any Date functions.
const date = new Date();
const ts = +date;
const mondayTS = ts - ts % (60 * 60 * 24 * (7-4) * 1000);
const monday = new Date(mondayTS);
console.log(monday.toISOString(), 'Day:', monday.getDay());
const formatTS = v => new Date(v).toISOString();
const adjust = (v, d = 1) => v - v % (d * 1000);
const d = new Date('2020-04-22T21:48:17.468Z');
const ts = +d; // 1587592097468
const test = v => console.log(formatTS(adjust(ts, v)));
test(); // 2020-04-22T21:48:17.000Z
test(60); // 2020-04-22T21:48:00.000Z
test(60 * 60); // 2020-04-22T21:00:00.000Z
test(60 * 60 * 24); // 2020-04-22T00:00:00.000Z
test(60 * 60 * 24 * (7-4)); // 2020-04-20T00:00:00.000Z, monday
// So, what does `(7-4)` mean?
// 7 - days number in the week
// 4 - shifting for the weekday number of the first second of the 1970 year, the first time stamp second.
// new Date(0) ---> 1970-01-01T00:00:00.000Z
// new Date(0).getDay() ---> 4

It is important to discern between local time and UTC. I wanted to find the start of the week in UTC, so I used the following function.
function start_of_week_utc(date, start_day = 1) {
// Returns the start of the week containing a 'date'. Monday 00:00 UTC is
// considered to be the boundary between adjacent weeks, unless 'start_day' is
// specified. A Date object is returned.
date = new Date(date);
const day_of_month = date.getUTCDate();
const day_of_week = date.getUTCDay();
const difference_in_days = (
day_of_week >= start_day
? day_of_week - start_day
: day_of_week - start_day + 7
);
date.setUTCDate(day_of_month - difference_in_days);
date.setUTCHours(0);
date.setUTCMinutes(0);
date.setUTCSeconds(0);
date.setUTCMilliseconds(0);
return date;
}
To find the start of the week in a given timezone, first add the timezone offset to the input date and then subtract it from the output date.
const local_start_of_week = new Date(
start_of_week_utc(
date.getTime() + timezone_offset_ms
).getTime() - timezone_offset_ms
);

I use this:
let current_date = new Date();
let days_to_monday = 1 - current_date.getDay();
monday_date = current_date.addDays(days_to_monday);
// https://stackoverflow.com/a/563442/6533037
Date.prototype.addDays = function(days) {
var date = new Date(this.valueOf());
date.setDate(date.getDate() + days);
return date;
}
It works fine.

Accepted answer won't work for anyone who runs the code in UTC-XX:XX timezone.
Here is code which will work regardless of timezone for date only. This won't work if you provide time too. Only provide date or parse date and provide it as input. I have mentioned different test cases at start of the code.
function getDateForTheMonday(dateString) {
var orignalDate = new Date(dateString)
var modifiedDate = new Date(dateString)
var day = modifiedDate.getDay()
diff = modifiedDate.getDate() - day + (day == 0 ? -6:1);// adjust when day is sunday
modifiedDate.setDate(diff)
var diffInDate = orignalDate.getDate() - modifiedDate.getDate()
if(diffInDate == 6) {
diff = diff + 7
modifiedDate.setDate(diff)
}
console.log("Given Date : " + orignalDate.toUTCString())
console.log("Modified date for Monday : " + modifiedDate)
}
getDateForTheMonday("2022-08-01") // Jul month with 31 Days
getDateForTheMonday("2022-07-01") // June month with 30 days
getDateForTheMonday("2022-03-01") // Non leap year February
getDateForTheMonday("2020-03-01") // Leap year February
getDateForTheMonday("2022-01-01") // First day of the year
getDateForTheMonday("2021-12-31") // Last day of the year

Extending answer from #Christian C. Salvadó and information from #Ayyash (object is mutable) and #Awi and #Louis Ameline (set hours to 00:00:00)
The function can be like this
function getMonday(d) {
var day = d.getDay(),
diff = d.getDate() - day + (day == 0 ? -6:1); // adjust when day is sunday
d.setDate(diff);
d.setHours(0,0,0,0); // set hours to 00:00:00
return d; // object is mutable no need to recreate object
}
getMonday(new Date())

Check out: moment.js
Example:
moment().day(-7); // last Sunday (0 - 7)
moment().day(7); // next Sunday (0 + 7)
moment().day(10); // next Wednesday (3 + 7)
moment().day(24); // 3 Wednesdays from now (3 + 7 + 7 + 7)
Bonus: works with node.js too

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.