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I am learning to use Date object in Javascript. Tried to calculate difference between now and some set date, but it returns a much larger value than inteded. The codepen is here, I can't seem to figure what I did wrong... Help?
var setdate = new Date(2014, 4, 27, 14,30); //27th of April this year at 14:30
var now = new Date(); //Now, whenever this code runs
var diff = Math.round((setdate.getTime() - now.getTime())/1000); //Difference in seconds
function NiceTimer(delta) { //Decompose the difference in seconds into date units.
this.days = Math.floor(delta/ 86400);
delta -= this.days*86400; //Subtract the value once it has been "extracted".
this.hours = Math.floor(delta/ 3600);
delta -= this.hours*3600;
this.minutes = Math.floor(delta/ 60);
delta -= this.minutes*60;
this.seconds = delta;
this.printString = function() {
return "The event starts in "+this.days+" days, "+this.hours+" hours, "+this.minutes+" minutes and "+this.seconds+" seconds"; //Output a readable countdown string
}
}
var timer = new NiceTimer(diff);
var el = document.getElementById("timer");
el.innerHTML = timer.printString();
var setdate = new Date(2014, 4, 27, 14,30); //27th of April this year at 14:30
Change the four to a three, months start at index zero.
var setdate = new Date(2014, 3, 27, 14,30);
Date # MDN:
month
Integer value representing the month, beginning with 0 for January to 11 for December.
How do I use JavaScript to calculate the day of the year, from 1 - 366?
For example:
January 3 should be 3.
February 1 should be 32.
Following OP's edit:
var now = new Date();
var start = new Date(now.getFullYear(), 0, 0);
var diff = now - start;
var oneDay = 1000 * 60 * 60 * 24;
var day = Math.floor(diff / oneDay);
console.log('Day of year: ' + day);
Edit: The code above will fail when now is a date in between march 26th and October 29th andnow's time is before 1AM (eg 00:59:59). This is due to the code not taking daylight savings time into account. You should compensate for this:
var now = new Date();
var start = new Date(now.getFullYear(), 0, 0);
var diff = (now - start) + ((start.getTimezoneOffset() - now.getTimezoneOffset()) * 60 * 1000);
var oneDay = 1000 * 60 * 60 * 24;
var day = Math.floor(diff / oneDay);
console.log('Day of year: ' + day);
I find it very interesting that no one considered using UTC since it is not subject to DST. Therefore, I propose the following:
function daysIntoYear(date){
return (Date.UTC(date.getFullYear(), date.getMonth(), date.getDate()) - Date.UTC(date.getFullYear(), 0, 0)) / 24 / 60 / 60 / 1000;
}
You can test it with the following:
[new Date(2016,0,1), new Date(2016,1,1), new Date(2016,2,1), new Date(2016,5,1), new Date(2016,11,31)]
.forEach(d =>
console.log(`${d.toLocaleDateString()} is ${daysIntoYear(d)} days into the year`));
Which outputs for the leap year 2016 (verified using http://www.epochconverter.com/days/2016):
1/1/2016 is 1 days into the year
2/1/2016 is 32 days into the year
3/1/2016 is 61 days into the year
6/1/2016 is 153 days into the year
12/31/2016 is 366 days into the year
This works across Daylight Savings Time changes in all countries (the "noon" one above doesn't work in Australia):
Date.prototype.isLeapYear = function() {
var year = this.getFullYear();
if((year & 3) != 0) return false;
return ((year % 100) != 0 || (year % 400) == 0);
};
// Get Day of Year
Date.prototype.getDOY = function() {
var dayCount = [0, 31, 59, 90, 120, 151, 181, 212, 243, 273, 304, 334];
var mn = this.getMonth();
var dn = this.getDate();
var dayOfYear = dayCount[mn] + dn;
if(mn > 1 && this.isLeapYear()) dayOfYear++;
return dayOfYear;
};
Date.prototype.dayOfYear= function(){
var j1= new Date(this);
j1.setMonth(0, 0);
return Math.round((this-j1)/8.64e7);
}
alert(new Date().dayOfYear())
Luckily this question doesn't specify if the number of the current day is required, leaving room for this answer.
Also some answers (also on other questions) had leap-year problems or used the Date-object. Although javascript's Date object covers approximately 285616 years (100,000,000 days) on either side of January 1 1970, I was fed up with all kinds of unexpected date inconsistencies across different browsers (most notably year 0 to 99). I was also curious how to calculate it.
So I wrote a simple and above all, small algorithm to calculate the correct (Proleptic Gregorian / Astronomical / ISO 8601:2004 (clause 4.3.2.1), so year 0 exists and is a leap year and negative years are supported) day of the year based on year, month and day.
Note that in AD/BC notation, year 0 AD/BC does not exist: instead year 1 BC is the leap-year! IF you need to account for BC notation then simply subtract one year of the (otherwise positive) year-value first!!
I modified (for javascript) the short-circuit bitmask-modulo leapYear algorithm and came up with a magic number to do a bit-wise lookup of offsets (that excludes jan and feb, thus needing 10 * 3 bits (30 bits is less than 31 bits, so we can safely save another character on the bitshift instead of >>>)).
Note that neither month or day may be 0. That means that if you need this equation just for the current day (feeding it using .getMonth()) you just need to remove the -- from --m.
Note this assumes a valid date (although error-checking is just some characters more).
function dayNo(y,m,d){
return --m*31-(m>1?(1054267675>>m*3-6&7)-(y&3||!(y%25)&&y&15?0:1):0)+d;
}
<!-- some examples for the snippet -->
<input type=text value="(-)Y-M-D" onblur="
var d=this.value.match(/(-?\d+)[^\d]+(\d\d?)[^\d]+(\d\d?)/)||[];
this.nextSibling.innerHTML=' Day: ' + dayNo(+d[1], +d[2], +d[3]);
" /><span></span>
<br><hr><br>
<button onclick="
var d=new Date();
this.nextSibling.innerHTML=dayNo(d.getFullYear(), d.getMonth()+1, d.getDate()) + ' Day(s)';
">get current dayno:</button><span></span>
Here is the version with correct range-validation.
function dayNo(y,m,d){
return --m>=0 && m<12 && d>0 && d<29+(
4*(y=y&3||!(y%25)&&y&15?0:1)+15662003>>m*2&3
) && m*31-(m>1?(1054267675>>m*3-6&7)-y:0)+d;
}
<!-- some examples for the snippet -->
<input type=text value="(-)Y-M-D" onblur="
var d=this.value.match(/(-?\d+)[^\d]+(\d\d?)[^\d]+(\d\d?)/)||[];
this.nextSibling.innerHTML=' Day: ' + dayNo(+d[1], +d[2], +d[3]);
" /><span></span>
Again, one line, but I split it into 3 lines for readability (and following explanation).
The last line is identical to the function above, however the (identical) leapYear algorithm is moved to a previous short-circuit section (before the day-number calculation), because it is also needed to know how much days a month has in a given (leap) year.
The middle line calculates the correct offset number (for max number of days) for a given month in a given (leap)year using another magic number: since 31-28=3 and 3 is just 2 bits, then 12*2=24 bits, we can store all 12 months. Since addition can be faster then subtraction, we add the offset (instead of subtract it from 31). To avoid a leap-year decision-branch for February, we modify that magic lookup-number on the fly.
That leaves us with the (pretty obvious) first line: it checks that month and date are within valid bounds and ensures us with a false return value on range error (note that this function also should not be able to return 0, because 1 jan 0000 is still day 1.), providing easy error-checking: if(r=dayNo(/*y, m, d*/)){}.
If used this way (where month and day may not be 0), then one can change --m>=0 && m<12 to m>0 && --m<12 (saving another char).
The reason I typed the snippet in it's current form is that for 0-based month values, one just needs to remove the -- from --m.
Extra:
Note, don't use this day's per month algorithm if you need just max day's per month. In that case there is a more efficient algorithm (because we only need leepYear when the month is February) I posted as answer this question: What is the best way to determine the number of days in a month with javascript?.
If used moment.js, we can get or even set the day of the year.
moment().dayOfYear();
//for getting
moment().dayOfYear(Number);
//for setting
moment.js is using this code for day of year calculation
If you don't want to re-invent the wheel, you can use the excellent date-fns (node.js) library:
var getDayOfYear = require('date-fns/get_day_of_year')
var dayOfYear = getDayOfYear(new Date(2017, 1, 1)) // 1st february => 32
This is my solution:
Math.floor((Date.now() - new Date(new Date().getFullYear(), 0, 0)) / 86400000)
Demo:
const getDateOfYear = (date) =>
Math.floor((date.getTime() - new Date(date.getFullYear(), 0, 0)) / 864e5);
const dayOfYear = getDateOfYear(new Date());
console.log(dayOfYear);
const dayOfYear = date => {
const myDate = new Date(date);
const year = myDate.getFullYear();
const firstJan = new Date(year, 0, 1);
const differenceInMillieSeconds = myDate - firstJan;
return (differenceInMillieSeconds / (1000 * 60 * 60 * 24) + 1);
};
const result = dayOfYear("2019-2-01");
console.log(result);
Well, if I understand you correctly, you want 366 on a leap year, 365 otherwise, right? A year is a leap year if it's evenly divisible by 4 but not by 100 unless it's also divisible by 400:
function daysInYear(year) {
if(year % 4 === 0 && (year % 100 !== 0 || year % 400 === 0)) {
// Leap year
return 366;
} else {
// Not a leap year
return 365;
}
}
Edit after update:
In that case, I don't think there's a built-in method; you'll need to do this:
function daysInFebruary(year) {
if(year % 4 === 0 && (year % 100 !== 0 || year % 400 === 0)) {
// Leap year
return 29;
} else {
// Not a leap year
return 28;
}
}
function dateToDay(date) {
var feb = daysInFebruary(date.getFullYear());
var aggregateMonths = [0, // January
31, // February
31 + feb, // March
31 + feb + 31, // April
31 + feb + 31 + 30, // May
31 + feb + 31 + 30 + 31, // June
31 + feb + 31 + 30 + 31 + 30, // July
31 + feb + 31 + 30 + 31 + 30 + 31, // August
31 + feb + 31 + 30 + 31 + 30 + 31 + 31, // September
31 + feb + 31 + 30 + 31 + 30 + 31 + 31 + 30, // October
31 + feb + 31 + 30 + 31 + 30 + 31 + 31 + 30 + 31, // November
31 + feb + 31 + 30 + 31 + 30 + 31 + 31 + 30 + 31 + 30, // December
];
return aggregateMonths[date.getMonth()] + date.getDate();
}
(Yes, I actually did that without copying or pasting. If there's an easy way I'll be mad)
This is a simple way to find the current day in the year, and it should account for leap years without a problem:
Javascript:
Math.round((new Date().setHours(23) - new Date(new Date().getYear()+1900, 0, 1, 0, 0, 0))/1000/60/60/24);
Javascript in Google Apps Script:
Math.round((new Date().setHours(23) - new Date(new Date().getYear(), 0, 1, 0, 0, 0))/1000/60/60/24);
The primary action of this code is to find the number of milliseconds that have elapsed in the current year and then convert this number into days. The number of milliseconds that have elapsed in the current year can be found by subtracting the number of milliseconds of the first second of the first day of the current year, which is obtained with new Date(new Date().getYear()+1900, 0, 1, 0, 0, 0) (Javascript) or new Date(new Date().getYear(), 0, 1, 0, 0, 0) (Google Apps Script), from the milliseconds of the 23rd hour of the current day, which was found with new Date().setHours(23). The purpose of setting the current date to the 23rd hour is to ensure that the day of year is rounded correctly by Math.round().
Once you have the number of milliseconds of the current year, then you can convert this time into days by dividing by 1000 to convert milliseconds to seconds, then dividing by 60 to convert seconds to minutes, then dividing by 60 to convert minutes to hours, and finally dividing by 24 to convert hours to days.
Note: This post was edited to account for differences between JavaScript and JavaScript implemented in Google Apps Script. Also, more context was added for the answer.
I think this is more straightforward:
var date365 = 0;
var currentDate = new Date();
var currentYear = currentDate.getFullYear();
var currentMonth = currentDate.getMonth();
var currentDay = currentDate.getDate();
var monthLength = [31,28,31,30,31,30,31,31,30,31,30,31];
var leapYear = new Date(currentYear, 1, 29);
if (leapYear.getDate() == 29) { // If it's a leap year, changes 28 to 29
monthLength[1] = 29;
}
for ( i=0; i < currentMonth; i++ ) {
date365 = date365 + monthLength[i];
}
date365 = date365 + currentDay; // Done!
This method takes into account timezone issue and daylight saving time
function dayofyear(d) { // d is a Date object
var yn = d.getFullYear();
var mn = d.getMonth();
var dn = d.getDate();
var d1 = new Date(yn,0,1,12,0,0); // noon on Jan. 1
var d2 = new Date(yn,mn,dn,12,0,0); // noon on input date
var ddiff = Math.round((d2-d1)/864e5);
return ddiff+1;
}
(took from here)
See also this fiddle
Math.round((new Date().setHours(23) - new Date(new Date().getFullYear(), 0, 1, 0, 0, 0))/1000/86400);
further optimizes the answer.
Moreover, by changing setHours(23) or the last-but-two zero later on to another value may provide day-of-year related to another timezone.
For example, to retrieve from Europe a resource located in America.
This might be useful to those who need the day of the year as a string and have jQuery UI available.
You can use jQuery UI Datepicker:
day_of_year_string = $.datepicker.formatDate("o", new Date())
Underneath it works the same way as some of the answers already mentioned ((date_ms - first_date_of_year_ms) / ms_per_day):
function getDayOfTheYearFromDate(d) {
return Math.round((new Date(d.getFullYear(), d.getMonth(), d.getDate()).getTime()
- new Date(d.getFullYear(), 0, 0).getTime()) / 86400000);
}
day_of_year_int = getDayOfTheYearFromDate(new Date())
maybe help anybody
let day = (date => {
return Math.floor((date - new Date(date.getFullYear(), 0, 0)) / 1000 / 60 / 60 / 24)
})(new Date())
I've made one that's readable and will do the trick very quickly, as well as handle JS Date objects with disparate time zones.
I've included quite a few test cases for time zones, DST, leap seconds and Leap years.
P.S. ECMA-262 ignores leap seconds, unlike UTC. If you were to convert this to a language that uses real UTC, you could just add 1 to oneDay.
// returns 1 - 366
findDayOfYear = function (date) {
var oneDay = 1000 * 60 * 60 * 24; // A day in milliseconds
var og = { // Saving original data
ts: date.getTime(),
dom: date.getDate(), // We don't need to save hours/minutes because DST is never at 12am.
month: date.getMonth()
}
date.setDate(1); // Sets Date of the Month to the 1st.
date.setMonth(0); // Months are zero based in JS's Date object
var start_ts = date.getTime(); // New Year's Midnight JS Timestamp
var diff = og.ts - start_ts;
date.setDate(og.dom); // Revert back to original date object
date.setMonth(og.month); // This method does preserve timezone
return Math.round(diff / oneDay) + 1; // Deals with DST globally. Ceil fails in Australia. Floor Fails in US.
}
// Tests
var pre_start_dst = new Date(2016, 2, 12);
var on_start_dst = new Date(2016, 2, 13);
var post_start_dst = new Date(2016, 2, 14);
var pre_end_dst_date = new Date(2016, 10, 5);
var on_end_dst_date = new Date(2016, 10, 6);
var post_end_dst_date = new Date(2016, 10, 7);
var pre_leap_second = new Date(2015, 5, 29);
var on_leap_second = new Date(2015, 5, 30);
var post_leap_second = new Date(2015, 6, 1);
// 2012 was a leap year with a leap second in june 30th
var leap_second_december31_premidnight = new Date(2012, 11, 31, 23, 59, 59, 999);
var january1 = new Date(2016, 0, 1);
var january31 = new Date(2016, 0, 31);
var december31 = new Date(2015, 11, 31);
var leap_december31 = new Date(2016, 11, 31);
alert( ""
+ "\nPre Start DST: " + findDayOfYear(pre_start_dst) + " === 72"
+ "\nOn Start DST: " + findDayOfYear(on_start_dst) + " === 73"
+ "\nPost Start DST: " + findDayOfYear(post_start_dst) + " === 74"
+ "\nPre Leap Second: " + findDayOfYear(pre_leap_second) + " === 180"
+ "\nOn Leap Second: " + findDayOfYear(on_leap_second) + " === 181"
+ "\nPost Leap Second: " + findDayOfYear(post_leap_second) + " === 182"
+ "\nPre End DST: " + findDayOfYear(pre_end_dst_date) + " === 310"
+ "\nOn End DST: " + findDayOfYear(on_end_dst_date) + " === 311"
+ "\nPost End DST: " + findDayOfYear(post_end_dst_date) + " === 312"
+ "\nJanuary 1st: " + findDayOfYear(january1) + " === 1"
+ "\nJanuary 31st: " + findDayOfYear(january31) + " === 31"
+ "\nNormal December 31st: " + findDayOfYear(december31) + " === 365"
+ "\nLeap December 31st: " + findDayOfYear(leap_december31) + " === 366"
+ "\nLast Second of Double Leap: " + findDayOfYear(leap_second_december31_premidnight) + " === 366"
);
I would like to provide a solution that does calculations adding the days for each previous month:
function getDayOfYear(date) {
var month = date.getMonth();
var year = date.getFullYear();
var days = date.getDate();
for (var i = 0; i < month; i++) {
days += new Date(year, i+1, 0).getDate();
}
return days;
}
var input = new Date(2017, 7, 5);
console.log(input);
console.log(getDayOfYear(input));
This way you don't have to manage the details of leap years and daylight saving.
A alternative using UTC timestamps. Also as others noted the day indicating 1st a month is 1 rather than 0. The month starts at 0 however.
var now = Date.now();
var year = new Date().getUTCFullYear();
var year_start = Date.UTC(year, 0, 1);
var day_length_in_ms = 1000*60*60*24;
var day_number = Math.floor((now - year_start)/day_length_in_ms)
console.log("Day of year " + day_number);
You can pass parameter as date number in setDate function:
var targetDate = new Date();
targetDate.setDate(1);
// Now we can see the expected date as: Mon Jan 01 2018 01:43:24
console.log(targetDate);
targetDate.setDate(365);
// You can see: Mon Dec 31 2018 01:44:47
console.log(targetDate)
For those among us who want a fast alternative solution.
(function(){"use strict";
function daysIntoTheYear(dateInput){
var fullYear = dateInput.getFullYear()|0;
// "Leap Years are any year that can be exactly divided by 4 (2012, 2016, etc)
// except if it can be exactly divided by 100, then it isn't (2100, 2200, etc)
// except if it can be exactly divided by 400, then it is (2000, 2400)"
// (https://www.mathsisfun.com/leap-years.html).
var isLeapYear = ((fullYear & 3) | (fullYear/100 & 3)) === 0 ? 1 : 0;
// (fullYear & 3) = (fullYear % 4), but faster
//Alternative:var isLeapYear=(new Date(currentYear,1,29,12)).getDate()===29?1:0
var fullMonth = dateInput.getMonth()|0;
return ((
// Calculate the day of the year in the Gregorian calendar
// The code below works based upon the facts of signed right shifts
// • (x) >> n: shifts n and fills in the n highest bits with 0s
// • (-x) >> n: shifts n and fills in the n highest bits with 1s
// (This assumes that x is a positive integer)
(31 & ((-fullMonth) >> 4)) + // January // (-11)>>4 = -1
((28 + isLeapYear) & ((1-fullMonth) >> 4)) + // February
(31 & ((2-fullMonth) >> 4)) + // March
(30 & ((3-fullMonth) >> 4)) + // April
(31 & ((4-fullMonth) >> 4)) + // May
(30 & ((5-fullMonth) >> 4)) + // June
(31 & ((6-fullMonth) >> 4)) + // July
(31 & ((7-fullMonth) >> 4)) + // August
(30 & ((8-fullMonth) >> 4)) + // September
(31 & ((9-fullMonth) >> 4)) + // October
(30 & ((10-fullMonth) >> 4)) + // November
// There are no months past December: the year rolls into the next.
// Thus, fullMonth is 0-based, so it will never be 12 in Javascript
(dateInput.getDate()|0) // get day of the month
)&0xffff);
}
// Demonstration:
var date = new Date(2100, 0, 1)
for (var i=0; i<12; i=i+1|0, date.setMonth(date.getMonth()+1|0))
console.log(date.getMonth()+":\tday "+daysIntoTheYear(date)+"\t"+date);
date = new Date(1900, 0, 1);
for (var i=0; i<12; i=i+1|0, date.setMonth(date.getMonth()+1|0))
console.log(date.getMonth()+":\tday "+daysIntoTheYear(date)+"\t"+date);
// Performance Benchmark:
console.time("Speed of processing 65536 dates");
for (var i=0,month=date.getMonth()|0; i<65536; i=i+1|0)
date.setMonth(month=month+1+(daysIntoTheYear(date)|0)|0);
console.timeEnd("Speed of processing 65536 dates");
})();
The size of the months of the year and the way that Leap Years work fits perfectly into keeping our time on track with the sun. Heck, it works so perfectly that all we ever do is just adjust mere seconds here and there. Our current system of leap years has been in effect since February 24th, 1582, and will likely stay in effect for the foreseeable future.
DST, however, is very subject to change. It may be that 20 years from now, some country may offset time by a whole day or some other extreme for DST. A whole DST day will almost certainly never happen, but DST is still nevertheless very up-in-the-air and indecisive. Thus, the above solution is future proof in addition to being very very fast.
The above code snippet runs very fast. My computer can process 65536 dates in ~52ms on Chrome.
This is a solution that avoids the troublesome Date object and timezone issues, it requires that your input date be in the format "yyyy-dd-mm". If you want to change the format, then modify date_str_to_parts function:
function get_day_of_year(str_date){
var date_parts = date_str_to_parts(str_date);
var is_leap = (date_parts.year%4)==0;
var acct_for_leap = (is_leap && date_parts.month>2);
var day_of_year = 0;
var ary_months = [
0,
31, //jan
28, //feb(non leap)
31, //march
30, //april
31, //may
30, //june
31, //july
31, //aug
30, //sep
31, //oct
30, //nov
31 //dec
];
for(var i=1; i < date_parts.month; i++){
day_of_year += ary_months[i];
}
day_of_year += date_parts.date;
if( acct_for_leap ) day_of_year+=1;
return day_of_year;
}
function date_str_to_parts(str_date){
return {
"year":parseInt(str_date.substr(0,4),10),
"month":parseInt(str_date.substr(5,2),10),
"date":parseInt(str_date.substr(8,2),10)
}
}
A straightforward solution with complete explanation.
var dayOfYear = function(date) {
const daysInMonth = [31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31];
const [yyyy, mm, dd] = date.split('-').map(Number);
// Checks if February has 29 days
const isLeap = (year) => new Date(year, 1, 29).getDate() === 29;
// If it's a leap year, changes 28 to 29
if (isLeap(yyyy)) daysInMonth[1] = 29;
let daysBeforeMonth = 0;
// Slice the array and exclude the current Month
for (const i of daysInMonth.slice(0, mm - 1)) {
daysBeforeMonth += i;
}
return daysBeforeMonth + dd;
};
console.log(dayOfYear('2020-1-3'));
console.log(dayOfYear('2020-2-1'));
I wrote these two javascript functions which return the day of the year (Jan 1 = 1).
Both of them account for leap years.
function dayOfTheYear() {
// for today
var M=[31,28,31,30,31,30,31,31,30,31,30,31]; var x=new Date(); var m=x.getMonth();
var y=x.getFullYear(); if (y % 400 == 0 || (y % 4 == 0 && y % 100 != 0)) {++M[1];}
var Y=0; for (var i=0;i<m;++i) {Y+=M[i];}
return Y+x.getDate();
}
function dayOfTheYear2(m,d,y) {
// for any day : m is 1 to 12, d is 1 to 31, y is a 4-digit year
var m,d,y; var M=[31,28,31,30,31,30,31,31,30,31,30,31];
if (y % 400 == 0 || (y % 4 == 0 && y % 100 != 0)) {++M[1];}
var Y=0; for (var i=0;i<m-1;++i) {Y+=M[i];}
return Y+d;
}
One Line:
Array.from(new Array(new Date().getMonth()), (x, i) => i).reduce((c, p, idx, array)=>{
let returnValue = c + new Date(new Date().getFullYear(), p, 0).getDate();
if(idx == array.length -1){
returnValue = returnValue + new Date().getDate();
}
return returnValue;
}, 0)
I needed a reliable (leap year and time zone resistant) algorithm for an application that makes heavy use of this feature, I found some algorithm written in the 90s and found that there is still no such efficient and stable solution here:
function dayOfYear1 (date) {
const year = date.getFullYear();
const month = date.getMonth()+1;
const day = date.getDate();
const N1 = Math.floor(275 * month / 9);
const N2 = Math.floor((month + 9) / 12);
const N3 = (1 + Math.floor((year - 4 * Math.floor(year / 4) + 2) / 3));
const N = N1 - (N2 * N3) + day - 30;
return N;
}
Algorithm works correctly in leap years, it does not depend on time zones with Date() and on top of that it is more efficient than any of the lower ones:
function dayOfYear2 (date) {
const monthsDays = [0, 31, 59, 90, 120, 151, 181, 212, 243, 273, 304, 334];
const year = date.getFullYear();
const month = date.getMonth();
const day = date.getDate();
let N = monthsDays[month] + day;
if ( month>1 && year%4==0 )
N++;
return N;
}
function dayOfYear3 (date) {
const yearDate = new Date(date.getFullYear(), 0, 0);
const timeZoneDiff = yearDate.getTimezoneOffset() - date.getTimezoneOffset();
const N = Math.floor(((date - yearDate )/1000/60 + timeZoneDiff)/60/24);
return N;
}
All of them are correct and work under the conditions mentioned above.
Performance comparison in 100k loop:
dayOfYear1 - 15 ms
dayOfYear2 - 17 ms
dayOfYear3 - 80 ms
It always get's me worried when mixing maths with date functions (it's so easy to miss some leap year other detail). Say you have:
var d = new Date();
I would suggest using the following, days will be saved in day:
for(var day = d.getDate(); d.getMonth(); day += d.getDate())
d.setDate(0);
Can't see any reason why this wouldn't work just fine (and I wouldn't be so worried about the few iterations since this will not be used so intensively).
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
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.
I want to find the number of the week which a day belongs to in javascript. My definition of week is the number of saturdays there have been up to that date. So the first few days of january are counted as the final week of the previous year. From there, the first saturday to the preceding friday is week 0 (0 based week counting).
Anyone know how to implement a function like this?
This?
function week(date) {
var firstSat = new Date(date.getFullYear(), 0, 1);
firstSat.setDate(firstSat.getDate() + (6 - firstSat.getDay()));
var delta = Math.floor((date - firstSat) / (7*24*60*60*1000));
return delta < 0 ?
delta + 52 : // first few days before 1st Sat
delta
}
week(new Date(2009,0,1)); // Jan 1, 2009 => 51 => "1 week into last year"
week(new Date(2009,0,2)); // 51
week(new Date(2009,0,3)); // 0 => "[beginning of] week 1"
week(new Date(2009,0,10)); // 1 => "[beginning of] week 2"
week(new Date(2009,0,11)); // 1 => "still week 2"
week(new Date(2009, 9, 30)); // Fri Oct 30, 2009 => 42
Checkout http://www.datejs.com/
var getWeekForDate = function(date) {
var yearStart = new Date(date.getFullYear(), 0, 0);
// move to first Saturday
yearStart.setDate(yearStart.getDate() + (6 - yearStart.getDay()));
var msSinceStart = date - yearStart;
var daysSinceStart = msSinceStart / (1000 * 60 * 60 * 24);
var weeksSinceStart = daysSinceStart / 7;
return Math.floor(weeksSinceStart);
};
In the end, I went for:
Date.prototype.getWeek = function()
{
var localOffset = this.getTimezoneOffset() * 60000;
var date = new Date(this-localOffset);
var firstSat = new Date(this);
firstSat.setDate(1);
firstSat.setMonth(0);
firstSat.setDate(firstSat.getDate() + (6 - firstSat.getDay()));
if(date < firstSat)
{
date.setDate(date.getDate()-7);
return date.getWeek()+1;
}
else
{
return Math.floor((date - firstSat) / (7*24*60*60*1000));
}
}
Which removes problems with UTC offsets