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How to calculate days between two date excluding weekends in angularjs? [duplicate]

This question already has answers here:
Find day difference between two dates (excluding weekend days)
(13 answers)
Closed 3 years ago.
In my application i have two date picker as start date and end date. when user choose start and end date the system will show the days between two dates but excluding the saturday and sunday. How to calculate it by using angularjs?

Does something like this work:
var startDate = new Date("01-10-2020");
var endDate = new Date("01-20-2020");
var nextDay = new Date(startDate);
var cnt = 0;
do {
/*if (nextDay.getDay() >= 1 && nextDay.getDay() <= 5) {
cnt = cnt + 1;
}*/
cnt += (nextDay.getDay() >= 1 && nextDay.getDay() <= 5) ? 1 : 0;
nextDay.setDate(nextDay.getDate() + 1);
} while (nextDay <= endDate);
console.log("Number of week days between " + startDate + " and " + endDate + " = " + cnt);
Here is the fiddler.

You don't want to do an expensive loop over every day to see whether it is Saturday or Sunday. The logic should be as follows:
Work in UTC so we don't need to worry about time zone
Calculate total number of calendar weeks. In a single calendar week, there are guaranteed to be 5 weekdays (not weekends == SAT || SUN)
Calculate the remainder of days. This will be added to the calculation later.
Determine the "finalAdjustment" by seeing if the remainder falls on weekend days.
The number of weekdays is (5 * numWeeks) + remainderDays + finalAdjust
(function() {
"use strict";
var SUN = 0;
var MON = 1;
var TUE = 2;
var WED = 3;
var THU = 4;
var FRI = 5;
var SAT = 6;
function isWeekendDay(day) {
return day === SAT || day === SUN;
}
function numberWeekDays(start, end) {
var numCalendarDays = (end - start) / 1000 / 60 / 60 / 24;
var numWeeks = Math.floor(numCalendarDays / 7);
// Potential days to add on to the number of full calendar
// weeks. This will be adjusted by "finalAdjust"
var remainderDays = numCalendarDays % 7;
// Adjustments for start and end dates being on a weekend
// ----------------------------
// Start at one because the same day should count as 1
// but number of days between same day is 0 based on
// arithmetic above.
// Change this to 0 if you don't want end date inclusive...
var finalAdjust = 1;
var startDay = start.getUTCDay();
var endDay = end.getUTCDay();
// On a weekend, so adjust by subtracting 1
if (isWeekendDay(startDay)) {
finalAdjust--;
}
// On a weekend, so adjust by subtracting 1
if (isWeekendDay(endDay)) {
finalAdjust--;
}
// This accounts for subtracting an extra weekend when starting
// at the beginning of a weekend (e.g. Saturday into Monday)
// The end day cannot also be on a weekend based on week modular division (mod 7)
if (startDay === SAT && remainderDays > 2) {
finalAdjust--;
}
// ---------------------------
// For every full calendar week there are 5 week days
// Use that number with the remainderDays and finalAdjust above
// to arrive at the answer.
var numWeekDays = (5 * numWeeks) + remainderDays + finalAdjust;
return numWeekDays;
}
// Test cases
// Assume that the start and end dates are inclusive
// 2020-01-01 to 2020-01-01 is one day
// 2020-01-01 to 2020-01-02 is two days
// ----------------------
// A Wednesdday
var start = new Date("2020-01-08");
// A Saturday
var end = new Date("2020-02-01");
// Expected answer: 18
console.log(numberWeekDays(start, end));
// A Saturday
start = new Date("2020-01-05");
// A Monday
end = new Date("2020-01-31");
// Expected answer: 20
console.log(numberWeekDays(start, end));
// Weekday to weekday Tuesday to
start = new Date("2020-01-07");
end = new Date("2020-01-16");
// Expected: 8
console.log(numberWeekDays(start, end));
// Same week: Mon-Wed
start = new Date("2020-01-06");
end = new Date("2020-01-08");
// Expected answer: 3
console.log(numberWeekDays(start, end));
// Same day
start = new Date("2020-01-08");
end = new Date("2020-01-08");
// Expect: 1
console.log(numberWeekDays(start, end));
// Weekend only
start = new Date("2020-01-04");
end = new Date("2020-01-05");
// Expect: 0;
console.log(numberWeekDays(start, end));
// ------------------
}());
As others have stated, a date library like moment is useful here because it gives you a lot of utility functions for working with dates and durations.

exclude weekends in javascript date calculation

I have two sets of codes that work. Needed help combining them into one.
This code gets me the difference between two dates. works perfectly:
function test(){
var date1 = new Date(txtbox_1.value);
var date2 = new Date(txtbox_2.value);
var diff = (date2 - date1)/1000;
var diff = Math.abs(Math.floor(diff));
var days = Math.floor(diff/(24*60*60));
var leftSec = diff - days * 24*60*60;
var hrs = Math.floor(leftSec/(60*60));
var leftSec = leftSec - hrs * 60*60;
var min = Math.floor(leftSec/(60));
var leftSec = leftSec - min * 60;
txtbox_3.value = days + "." + hrs; }
source for the above code
The code below by #cyberfly appears to have the answer of excluding sat and sun which is what i needed. source. However, its in jquery and the above code is in JS. Therefore, needed help combining as i lacked that knowledge :(
<script type="text/javascript">
$("#startdate, #enddate").change(function() {
var d1 = $("#startdate").val();
var d2 = $("#enddate").val();
var minutes = 1000*60;
var hours = minutes*60;
var day = hours*24;
var startdate1 = getDateFromFormat(d1, "d-m-y");
var enddate1 = getDateFromFormat(d2, "d-m-y");
var days = calcBusinessDays(new Date(startdate1),new Date(enddate1));
if(days>0)
{ $("#noofdays").val(days);}
else
{ $("#noofdays").val(0);}
});
</script>
EDIT
Made an attempt at combining the codes. here is my sample. getting object expected error.
function test(){
var date1 = new Date(startdate.value);
var date2 = new Date(enddate.value);
var diff = (date2 - date1)/1000;
var diff = Math.abs(Math.floor(diff));
var days = Math.floor(diff/(24*60*60));
var leftSec = diff - days * 24*60*60;
var hrs = Math.floor(leftSec/(60*60));
var leftSec = leftSec - hrs * 60*60;
var min = Math.floor(leftSec/(60));
var leftSec = leftSec - min * 60;
var startdate1 = getDateFromFormat(startdate, "dd/mm/yyyy hh:mm");
var enddate1 = getDateFromFormat(enddate, "dd/mm/yyyy hh:mm");
days = calcBusinessDays(new Date(startdate1),new Date(enddate1));
noofdays.value = days + "." + hrs; }
start: <input type="text" id="startdate" name="startdate" value="02/03/2015 00:00">
end: <input type="text" id="enddate" name="enddate" value="02/03/2015 00:01">
<input type="text" id="noofdays" name="noofdays" value="">

When determining the number of days between two dates, there are lots of decisions to be made about what is a day. For example, the period 1 Feb to 2 Feb is generally one day, so 1 Feb to 1 Feb is zero days.
When adding the complexity of counting only business days, things get a lot tougher. E.g. Monday 2 Feb 2015 to Friday 6 February is 4 elapsed days (Monday to Tuesday is 1, Monday to Wednesday is 2, etc.), however the expression "Monday to Friday" is generally viewed as 5 business days and the duration Mon 2 Feb to Sat 7 Feb should also be 4 business days, but Sunday to Saturday should be 5.
So here's my algorithm:
Get the total number of whole days between the two dates
Divide by 7 to get the number of whole weeks
Multiply the number of weeks by two to get the number of weekend days
Subtract the number of weekend days from the whole to get business days
If the number of total days is not an even number of weeks, add the numbe of weeks * 7 to the start date to get a temp date
While the temp date is less than the end date:
if the temp date is not a Saturday or Sunday, add one the business days
add one to the temp date
That's it.
The stepping part at the end can probably be replaced by some other algorithm, but it will never loop for more than 6 days so it's a simple and reasonably efficient solution to the issue of uneven weeks.
Some consequences of the above:
Monday to Friday is 4 business days
Any day to the same day in a different week is an even number of weeks and therefore an even mutiple of 5, e.g. Monday 2 Feb to Monday 9 Feb and Sunday 1 Feb to Sunday 8 Feb are 5 business days
Friday 6 Feb to Sunday 7 Feb is zero business days
Friday 6 Feb to Monday 9 Feb is one business day
Sunday 8 Feb to: Sunday 15 Feb, Sat 14 Feb and Fri 13 Feb are all 5 business days
Here's the code:
// Expects start date to be before end date
// start and end are Date objects
function dateDifference(start, end) {
// Copy date objects so don't modify originals
var s = new Date(+start);
var e = new Date(+end);
// Set time to midday to avoid dalight saving and browser quirks
s.setHours(12,0,0,0);
e.setHours(12,0,0,0);
// Get the difference in whole days
var totalDays = Math.round((e - s) / 8.64e7);
// Get the difference in whole weeks
var wholeWeeks = totalDays / 7 | 0;
// Estimate business days as number of whole weeks * 5
var days = wholeWeeks * 5;
// If not even number of weeks, calc remaining weekend days
if (totalDays % 7) {
s.setDate(s.getDate() + wholeWeeks * 7);
while (s < e) {
s.setDate(s.getDate() + 1);
// If day isn't a Sunday or Saturday, add to business days
if (s.getDay() != 0 && s.getDay() != 6) {
++days;
}
}
}
return days;
}
I don't know how it compares to jfriend00's answer or the code you referenced, if you want the period to be inclusive, just add one if the start or end date are a business day.

Here's a simple function to calculate the number of business days between two date objects. As designed, it does not count the start day, but does count the end day so if you give it a date on a Tuesday of one week and a Tuesday of the next week, it will return 5 business days. This does not account for holidays, but does work properly across daylight savings changes.
function calcBusinessDays(start, end) {
// This makes no effort to account for holidays
// Counts end day, does not count start day
// make copies we can normalize without changing passed in objects
var start = new Date(start);
var end = new Date(end);
// initial total
var totalBusinessDays = 0;
// normalize both start and end to beginning of the day
start.setHours(0,0,0,0);
end.setHours(0,0,0,0);
var current = new Date(start);
current.setDate(current.getDate() + 1);
var day;
// loop through each day, checking
while (current <= end) {
day = current.getDay();
if (day >= 1 && day <= 5) {
++totalBusinessDays;
}
current.setDate(current.getDate() + 1);
}
return totalBusinessDays;
}
And, the jQuery + jQueryUI code for a demo:
// make both input fields into date pickers
$("#startDate, #endDate").datepicker();
// process click to calculate the difference between the two days
$("#calc").click(function(e) {
var diff = calcBusinessDays(
$("#startDate").datepicker("getDate"),
$("#endDate").datepicker("getDate")
);
$("#diff").html(diff);
});
And, here's a simple demo built with the date picker in jQueryUI: http://jsfiddle.net/jfriend00/z1txs10d/

const firstDate = new Date("December 30, 2020");
const secondDate = new Date("January 4, 2021");
const daysWithOutWeekEnd = [];
for (var currentDate = new Date(firstDate); currentDate <= secondDate; currentDate.setDate(currentDate.getDate() + 1)) {
// console.log(currentDate);
if (currentDate.getDay() != 0 && currentDate.getDay() != 6) {
daysWithOutWeekEnd.push(new Date(currentDate));
}
}
console.log(daysWithOutWeekEnd, daysWithOutWeekEnd.length);

#RobG has given an excellent algorithm to separate business days from weekends.
I think the only problem is if the starting days is a weekend, Saturday or Sunday, then the no of working days/weekends will one less.
Corrected code is below.
function dateDifference(start, end) {
// Copy date objects so don't modify originals
var s = new Date(start);
var e = new Date(end);
var addOneMoreDay = 0;
if( s.getDay() == 0 || s.getDay() == 6 ) {
addOneMoreDay = 1;
}
// Set time to midday to avoid dalight saving and browser quirks
s.setHours(12,0,0,0);
e.setHours(12,0,0,0);
// Get the difference in whole days
var totalDays = Math.round((e - s) / 8.64e7);
// Get the difference in whole weeks
var wholeWeeks = totalDays / 7 | 0;
// Estimate business days as number of whole weeks * 5
var days = wholeWeeks * 5;
// If not even number of weeks, calc remaining weekend days
if (totalDays % 7) {
s.setDate(s.getDate() + wholeWeeks * 7);
while (s < e) {
s.setDate(s.getDate() + 1);
// If day isn't a Sunday or Saturday, add to business days
if (s.getDay() != 0 && s.getDay() != 6) {
++days;
}
//s.setDate(s.getDate() + 1);
}
}
var weekEndDays = totalDays - days + addOneMoreDay;
return weekEndDays;
}
JSFiddle link is https://jsfiddle.net/ykxj4k09/2/

First Get the Number of Days in a month
totalDays(month, year) {
return new Date(year, month, 0).getDate();
}
Then Get No Of Working Days In A Month By removing Saturday and Sunday
totalWorkdays() {
var d = new Date(); // to know present date
var m = d.getMonth() + 1; // to know present month
var y = d.getFullYear(); // to knoow present year
var td = this.totalDays(m, y);// to get no of days in a month
for (var i = 1; i <= td; i++) {
var s = new Date(y, m - 1, i);
if (s.getDay() != 0 && s.getDay() != 6) {
this.workDays.push(s.getDate());// working days
}else {
this.totalWeekDays.push(s.getDate());//week days
}
}
this.totalWorkingDays = this.workDays.length;
}

I thought the above code snippets others shared are lengthy.
I am sharing a concise snippet that gives date after considering the total number of days specified. we can also customize dates other than Saturdays and Sundays.
function getBusinessDays(dateObj, days) {
for (var i = 0; i < days; i++) {
if (days > 0) {
switch (dateObj.getDay()) {
// 6 being Saturday and 0 being Sunday.
case 6, 0:
dateObj.setDate(dateObj.getDate() + 2)
break;
//5 = Friday.
case 5:
dateObj.setDate(dateObj.getDate() + 3)
break;
//handle Monday, Tuesday, Wednesday and Thursday!
default:
dateObj.setDate(dateObj.getDate() + 1)
//console.log(dateObj)
break;
}
}
}
return dateObj;
}
console.log(getBusinessDays(new Date(), 11))
//Mon Dec 20 2021 18:56:01 GMT+0530 (India Standard Time)

How to count month and days between two dates through JavaScript?

We have using below code.
var oneDay = 24*60*60*1000; // hours*minutes*seconds*milliseconds
var firstDate = new Date(2008,01,12);
var secondDate = new Date(2008,01,22);
var diffDays = Math.abs((firstDate.getTime() - secondDate.getTime())/(oneDay));
But we can't find both month and days between them.
I want like suppose start date is 27-sep-2014 and end date is 1-nov-2014. Result will be 1 month and 5 days.

There multiple ways to count months and days. Without further restrictions all correct of them are correct even though the results might differ.
For example from September 27 to November 1 could be either
1 month and 5 days or
4 days and 1 month.
This is one possible solution.
// swap dates if difference would be negative
if (firstDate.getTime() > secondDate.getTime()) {
var tmp = firstDate;
firstDate = secondDate;
secondDate = tmp;
}
var years = secondDate.getFullYear() - firstDate.getFullYear();
var months = secondDate.getMonth() - firstDate.getMonth();
var days = secondDate.getDate() - firstDate.getDate();
// prevent negative amount of days by breaking up months
for (var i = 0; days < 0; ++i) {
// while the day difference is negative
// we break up months into days, starting with the first
months -= 1;
days += new Date(
firstDate.getFullYear(),
firstDate.getMonth() + 1 + i,
0, 0, 0, 0, 0
).getDate();
}
// prevent negative amount of months by breaking up years
if (months < 0) {
years += Math.floor(months / 12);
months = (months % 12 + 12) % 12;
}
// print the result
console.log([
{amount: days, unit: 'day'},
{amount: months, unit: 'month'},
{amount: years, unit: 'year'},
].filter(value => value.amount).map(value =>
value.amount === 1 ?
`${value.amount} ${value.unit}` :
`${value.amount} ${value.unit}s`
).reduce((result, part, index, parts) =>
index > 0 ? index === parts.length - 1 ?
`${result} and ${part}` :
`${result}, ${part}` :
`${part}`,
`0 days`
));
Examples:
02/12 to 02/22: 10 days
09/27 to 11/01: 4 days and 1 month // instead of 1 month and 5 days
12/31 to 03/01: 1 day and 2 months // instead of 1 month and 29 days
05/31 to 06/30: 30 days
01/31 to 03/30: 30 days and 1 month // instead of 1 month and 27 days
10/27/2010 to 08/26/2014: 30 days, 9 months and 3 years

Try this
var oneDay = 24*60*60*1000;
var firstDate = new Date(2007,01,12);
var secondDate = new Date(2008,01,22);
var diffDays = Math.abs((firstDate.getTime() - secondDate.getTime()));
var result = '',
years, months, days;
if((years = diffDays / (365 * oneDay)) > 1){
result += Math.floor(years) + ' Year(s)';
diffDays %= (365 * oneDay);
}
if((months = diffDays / (30 * oneDay)) > 1){
result += Math.floor(months) + ' Month(s)';
diffDays %= (30 * oneDay);
}
result += (diffDays / oneDay) + ' Days(s)';
alert(result);

JavaScript Date.getWeek()? [duplicate]

This question already has answers here:
Get week of year in JavaScript like in PHP
(23 answers)
Closed 5 years ago.
I'm looking for a tested solid solution for getting current week of the year for specified date. All I can find are the ones that doesn't take in account leap years or just plain wrong. Does anyone have this type of stuff?
Or even better a function that says how many weeks does month occupy. It is usually 5, but can be 4 (feb) or 6 (1st is sunday and month has 30-31 days in it)
=================
UPDATE:
Still not sure about getting week #, but since I figured out it won't solve my problem with calculating how many weeks month occupy, I abandoned it.
Here's a function to find out how many weeks exactly month occupy on the calendar:
getWeeksNum: function(year, month) {
var daysNum = 32 - new Date(year, month, 32).getDate(),
fDayO = new Date(year, month, 1).getDay(),
fDay = fDayO ? (fDayO - 1) : 6,
weeksNum = Math.ceil((daysNum + fDay) / 7);
return weeksNum;
}

/**
* Returns the week number for this date. dowOffset is the day of week the week
* "starts" on for your locale - it can be from 0 to 6. If dowOffset is 1 (Monday),
* the week returned is the ISO 8601 week number.
* #param int dowOffset
* #return int
*/
Date.prototype.getWeek = function (dowOffset) {
/*getWeek() was developed by Nick Baicoianu at MeanFreePath: http://www.meanfreepath.com */
dowOffset = typeof(dowOffset) == 'number' ? dowOffset : 0; //default dowOffset to zero
var newYear = new Date(this.getFullYear(),0,1);
var day = newYear.getDay() - dowOffset; //the day of week the year begins on
day = (day >= 0 ? day : day + 7);
var daynum = Math.floor((this.getTime() - newYear.getTime() -
(this.getTimezoneOffset()-newYear.getTimezoneOffset())*60000)/86400000) + 1;
var weeknum;
//if the year starts before the middle of a week
if(day < 4) {
weeknum = Math.floor((daynum+day-1)/7) + 1;
if(weeknum > 52) {
nYear = new Date(this.getFullYear() + 1,0,1);
nday = nYear.getDay() - dowOffset;
nday = nday >= 0 ? nday : nday + 7;
/*if the next year starts before the middle of
the week, it is week #1 of that year*/
weeknum = nday < 4 ? 1 : 53;
}
}
else {
weeknum = Math.floor((daynum+day-1)/7);
}
return weeknum;
};
Usage:
var mydate = new Date(2011,2,3); // month number starts from 0
// or like this
var mydate = new Date('March 3, 2011');
alert(mydate.getWeek());
Source

For those looking for a more simple approach;
Date.prototype.getWeek = function() {
var onejan = new Date(this.getFullYear(),0,1);
var today = new Date(this.getFullYear(),this.getMonth(),this.getDate());
var dayOfYear = ((today - onejan + 86400000)/86400000);
return Math.ceil(dayOfYear/7)
};
Use with:
var today = new Date();
var currentWeekNumber = today.getWeek();
console.log(currentWeekNumber);

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)

Get week number
Date.prototype.getWeek = function() {
var dt = new Date(this.getFullYear(),0,1);
return Math.ceil((((this - dt) / 86400000) + dt.getDay()+1)/7);
};
var myDate = new Date(2013, 3, 25); // 2013, 25 April
console.log(myDate.getWeek());

I know this is an old question, but maybe it helps:
http://weeknumber.net/how-to/javascript
// This script is released to the public domain and may be used, modified and
// distributed without restrictions. Attribution not necessary but appreciated.
// Source: https://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);
}
// Returns the four-digit year corresponding to the ISO week of the date.
Date.prototype.getWeekYear = function() {
var date = new Date(this.getTime());
date.setDate(date.getDate() + 3 - (date.getDay() + 6) % 7);
return date.getFullYear();
}

/*get the week number by following the norms of ISO 8601*/
function getWeek(dt){
var calc=function(o){
if(o.dtmin.getDay()!=1){
if(o.dtmin.getDay()<=4 && o.dtmin.getDay()!=0)o.w+=1;
o.dtmin.setDate((o.dtmin.getDay()==0)? 2 : 1+(7-o.dtmin.getDay())+1);
}
o.w+=Math.ceil((((o.dtmax.getTime()-o.dtmin.getTime())/(24*60*60*1000))+1)/7);
},getNbDaysInAMonth=function(year,month){
var nbdays=31;
for(var i=0;i<=3;i++){
nbdays=nbdays-i;
if((dtInst=new Date(year,month-1,nbdays)) && dtInst.getDate()==nbdays && (dtInst.getMonth()+1)==month && dtInst.getFullYear()==year)
break;
}
return nbdays;
};
if(dt.getMonth()+1==1 && dt.getDate()>=1 && dt.getDate()<=3 && (dt.getDay()>=5 || dt.getDay()==0)){
var pyData={"dtmin":new Date(dt.getFullYear()-1,0,1,0,0,0,0),"dtmax":new Date(dt.getFullYear()-1,11,getNbDaysInAMonth(dt.getFullYear()-1,12),0,0,0,0),"w":0};
calc(pyData);
return pyData.w;
}else{
var ayData={"dtmin":new Date(dt.getFullYear(),0,1,0,0,0,0),"dtmax":new Date(dt.getFullYear(),dt.getMonth(),dt.getDate(),0,0,0,0),"w":0},
nd12m=getNbDaysInAMonth(dt.getFullYear(),12);
if(dt.getMonth()==12 && dt.getDay()!=0 && dt.getDay()<=3 && nd12m-dt.getDate()<=3-dt.getDay())ayData.w=1;else calc(ayData);
return ayData.w;
}
}
alert(getWeek(new Date(2017,01-1,01)));

Difference between dates in JavaScript

How to find the difference between two dates?

By using the Date object and its milliseconds value, differences can be calculated:
var a = new Date(); // Current date now.
var b = new Date(2010, 0, 1, 0, 0, 0, 0); // Start of 2010.
var d = (b-a); // Difference in milliseconds.
You can get the number of seconds (as a integer/whole number) by dividing the milliseconds by 1000 to convert it to seconds then converting the result to an integer (this removes the fractional part representing the milliseconds):
var seconds = parseInt((b-a)/1000);
You could then get whole minutes by dividing seconds by 60 and converting it to an integer, then hours by dividing minutes by 60 and converting it to an integer, then longer time units in the same way. From this, a function to get the maximum whole amount of a time unit in the value of a lower unit and the remainder lower unit can be created:
function get_whole_values(base_value, time_fractions) {
time_data = [base_value];
for (i = 0; i < time_fractions.length; i++) {
time_data.push(parseInt(time_data[i]/time_fractions[i]));
time_data[i] = time_data[i] % time_fractions[i];
}; return time_data;
};
// Input parameters below: base value of 72000 milliseconds, time fractions are
// 1000 (amount of milliseconds in a second) and 60 (amount of seconds in a minute).
console.log(get_whole_values(72000, [1000, 60]));
// -> [0,12,1] # 0 whole milliseconds, 12 whole seconds, 1 whole minute.
If you're wondering what the input parameters provided above for the second Date object are, see their names below:
new Date(<year>, <month>, <day>, <hours>, <minutes>, <seconds>, <milliseconds>);
As noted in the comments of this solution, you don't necessarily need to provide all these values unless they're necessary for the date you wish to represent.

I have found this and it works fine for me:
Calculating the Difference between Two Known Dates
Unfortunately, calculating a date interval such as days, weeks, or months between two known dates is not as easy because you can't just add Date objects together. In order to use a Date object in any sort of calculation, we must first retrieve the Date's internal millisecond value, which is stored as a large integer. The function to do that is Date.getTime(). Once both Dates have been converted, subtracting the later one from the earlier one returns the difference in milliseconds. The desired interval can then be determined by dividing that number by the corresponding number of milliseconds. For instance, to obtain the number of days for a given number of milliseconds, we would divide by 86,400,000, the number of milliseconds in a day (1000 x 60 seconds x 60 minutes x 24 hours):
Date.daysBetween = function( date1, date2 ) {
//Get 1 day in milliseconds
var one_day=1000*60*60*24;
// Convert both dates to milliseconds
var date1_ms = date1.getTime();
var date2_ms = date2.getTime();
// Calculate the difference in milliseconds
var difference_ms = date2_ms - date1_ms;
// Convert back to days and return
return Math.round(difference_ms/one_day);
}
//Set the two dates
var y2k = new Date(2000, 0, 1);
var Jan1st2010 = new Date(y2k.getFullYear() + 10, y2k.getMonth(), y2k.getDate());
var today= new Date();
//displays 726
console.log( 'Days since '
+ Jan1st2010.toLocaleDateString() + ': '
+ Date.daysBetween(Jan1st2010, today));
The rounding is optional, depending on whether you want partial days or not.
Reference

If you are looking for a difference expressed as a combination of years, months, and days, I would suggest this function:
function interval(date1, date2) {
if (date1 > date2) { // swap
var result = interval(date2, date1);
result.years = -result.years;
result.months = -result.months;
result.days = -result.days;
result.hours = -result.hours;
return result;
}
result = {
years: date2.getYear() - date1.getYear(),
months: date2.getMonth() - date1.getMonth(),
days: date2.getDate() - date1.getDate(),
hours: date2.getHours() - date1.getHours()
};
if (result.hours < 0) {
result.days--;
result.hours += 24;
}
if (result.days < 0) {
result.months--;
// days = days left in date1's month,
// plus days that have passed in date2's month
var copy1 = new Date(date1.getTime());
copy1.setDate(32);
result.days = 32-date1.getDate()-copy1.getDate()+date2.getDate();
}
if (result.months < 0) {
result.years--;
result.months+=12;
}
return result;
}
// Be aware that the month argument is zero-based (January = 0)
var date1 = new Date(2015, 4-1, 6);
var date2 = new Date(2015, 5-1, 9);
document.write(JSON.stringify(interval(date1, date2)));
This solution will treat leap years (29 February) and month length differences in a way we would naturally do (I think).
So for example, the interval between 28 February 2015 and 28 March 2015 will be considered exactly one month, not 28 days. If both those days are in 2016, the difference will still be exactly one month, not 29 days.
Dates with exactly the same month and day, but different year, will always have a difference of an exact number of years. So the difference between 2015-03-01 and 2016-03-01 will be exactly 1 year, not 1 year and 1 day (because of counting 365 days as 1 year).

// This is for first date
first = new Date(2010, 03, 08, 15, 30, 10); // Get the first date epoch object
document.write((first.getTime())/1000); // get the actual epoch values
second = new Date(2012, 03, 08, 15, 30, 10); // Get the second date epoch object
document.write((second.getTime())/1000); // get the actual epoch values
diff= second - first ;
one_day_epoch = 24*60*60 ; // calculating one epoch
if ( diff/ one_day_epoch > 365 ) // check if it is exceeding regular calendar year
{
alert( 'date is exceeding one year');
}

This answer, based on another one (link at end), is about the difference between two dates.
You can see how it works because it's simple, also it includes splitting the difference into
units of time (a function that I made) and converting to UTC to stop time zone problems.
function date_units_diff(a, b, unit_amounts) {
var split_to_whole_units = function (milliseconds, unit_amounts) {
// unit_amounts = list/array of amounts of milliseconds in a
// second, seconds in a minute, etc., for example "[1000, 60]".
time_data = [milliseconds];
for (i = 0; i < unit_amounts.length; i++) {
time_data.push(parseInt(time_data[i] / unit_amounts[i]));
time_data[i] = time_data[i] % unit_amounts[i];
}; return time_data.reverse();
}; if (unit_amounts == undefined) {
unit_amounts = [1000, 60, 60, 24];
};
var utc_a = new Date(a.toUTCString());
var utc_b = new Date(b.toUTCString());
var diff = (utc_b - utc_a);
return split_to_whole_units(diff, unit_amounts);
}
// Example of use:
var d = date_units_diff(new Date(2010, 0, 1, 0, 0, 0, 0), new Date()).slice(0,-2);
document.write("In difference: 0 days, 1 hours, 2 minutes.".replace(
/0|1|2/g, function (x) {return String( d[Number(x)] );} ));
How my code above works
A date/time difference, as milliseconds, can be calculated using the Date object:
var a = new Date(); // Current date now.
var b = new Date(2010, 0, 1, 0, 0, 0, 0); // Start of 2010.
var utc_a = new Date(a.toUTCString());
var utc_b = new Date(b.toUTCString());
var diff = (utc_b - utc_a); // The difference as milliseconds.
Then to work out the number of seconds in that difference, divide it by 1000 to convert
milliseconds to seconds, then change the result to an integer (whole number) to remove
the milliseconds (fraction part of that decimal): var seconds = parseInt(diff/1000).
Also, I could get longer units of time using the same process, for example:
- (whole) minutes, dividing seconds by 60 and changing the result to an integer,
- hours, dividing minutes by 60 and changing the result to an integer.
I created a function for doing that process of splitting the difference into
whole units of time, named split_to_whole_units, with this demo:
console.log(split_to_whole_units(72000, [1000, 60]));
// -> [1,12,0] # 1 (whole) minute, 12 seconds, 0 milliseconds.
This answer is based on this other one.

You can also use it
export function diffDateAndToString(small: Date, big: Date) {
// To calculate the time difference of two dates
const Difference_In_Time = big.getTime() - small.getTime()
// To calculate the no. of days between two dates
const Days = Difference_In_Time / (1000 * 3600 * 24)
const Mins = Difference_In_Time / (60 * 1000)
const Hours = Mins / 60
const diffDate = new Date(Difference_In_Time)
console.log({ date: small, now: big, diffDate, Difference_In_Days: Days, Difference_In_Mins: Mins, Difference_In_Hours: Hours })
var result = ''
if (Mins < 60) {
result = Mins + 'm'
} else if (Hours < 24) result = diffDate.getMinutes() + 'h'
else result = Days + 'd'
return { result, Days, Mins, Hours }
}
results in { result: '30d', Days: 30, Mins: 43200, Hours: 720 }

Date.prototype.addDays = function(days) {
var dat = new Date(this.valueOf())
dat.setDate(dat.getDate() + days);
return dat;
}
function getDates(startDate, stopDate) {
var dateArray = new Array();
var currentDate = startDate;
while (currentDate <= stopDate) {
dateArray.push(currentDate);
currentDate = currentDate.addDays(1);
}
return dateArray;
}
var dateArray = getDates(new Date(), (new Date().addDays(7)));
for (i = 0; i < dateArray.length; i ++ ) {
// alert (dateArray[i]);
date=('0'+dateArray[i].getDate()).slice(-2);
month=('0' +(dateArray[i].getMonth()+1)).slice(-2);
year=dateArray[i].getFullYear();
alert(date+"-"+month+"-"+year );
}

var DateDiff = function(type, start, end) {
let // or var
years = end.getFullYear() - start.getFullYear(),
monthsStart = start.getMonth(),
monthsEnd = end.getMonth()
;
var returns = -1;
switch(type){
case 'm': case 'mm': case 'month': case 'months':
returns = ( ( ( years * 12 ) - ( 12 - monthsEnd ) ) + ( 12 - monthsStart ) );
break;
case 'y': case 'yy': case 'year': case 'years':
returns = years;
break;
case 'd': case 'dd': case 'day': case 'days':
returns = ( ( end - start ) / ( 1000 * 60 * 60 * 24 ) );
break;
}
return returns;
}
Usage
var qtMonths = DateDiff('mm', new Date('2015-05-05'), new Date());
var qtYears = DateDiff('yy', new Date('2015-05-05'), new Date());
var qtDays = DateDiff('dd', new Date('2015-05-05'), new Date());
OR
var qtMonths = DateDiff('m', new Date('2015-05-05'), new Date()); // m || y || d
var qtMonths = DateDiff('month', new Date('2015-05-05'), new Date()); // month || year || day
var qtMonths = DateDiff('months', new Date('2015-05-05'), new Date()); // months || years || days
...
var DateDiff = function (type, start, end) {
let // or var
years = end.getFullYear() - start.getFullYear(),
monthsStart = start.getMonth(),
monthsEnd = end.getMonth()
;
if(['m', 'mm', 'month', 'months'].includes(type)/*ES6*/)
return ( ( ( years * 12 ) - ( 12 - monthsEnd ) ) + ( 12 - monthsStart ) );
else if(['y', 'yy', 'year', 'years'].includes(type))
return years;
else if (['d', 'dd', 'day', 'days'].indexOf(type) !== -1/*EARLIER JAVASCRIPT VERSIONS*/)
return ( ( end - start ) / ( 1000 * 60 * 60 * 24 ) );
else
return -1;
}