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# johndyer/easter.js

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Calculate Easter in JavaScript
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 /** * Calculates Easter in the Gregorian/Western (Catholic and Protestant) calendar * based on the algorithm by Oudin (1940) from http://www.tondering.dk/claus/cal/easter.php * @returns {array} [int month, int day] */ function getEaster(year) { var f = Math.floor, // Golden Number - 1 G = year % 19, C = f(year / 100), // related to Epact H = (C - f(C / 4) - f((8 * C + 13)/25) + 19 * G + 15) % 30, // number of days from 21 March to the Paschal full moon I = H - f(H/28) * (1 - f(29/(H + 1)) * f((21-G)/11)), // weekday for the Paschal full moon J = (year + f(year / 4) + I + 2 - C + f(C / 4)) % 7, // number of days from 21 March to the Sunday on or before the Paschal full moon L = I - J, month = 3 + f((L + 40)/44), day = L + 28 - 31 * f(month / 4); return [month,day]; }

### DHFW commented May 29, 2019

Thank you!

For others: be aware that this returns the month as 1 = January, 2 = February and so on.

I subtracted the month with one to be able to convert it to a Javascript Date easily in which the month is index based (0 = January, 1 = February).

### varvino commented Feb 6, 2020

Thanks a lot for this

### gavinr commented Mar 10, 2020

I am using this here: Easter Dates
Thanks!

### sunsero commented Apr 11, 2020

Is it just me or is the function returning Easter Monday this year? Happy Easter, anyway!

### sunsero commented Apr 11, 2020

Don't think so because Easter Sunday is calculated for 2019 and for 2021 correctly. If it matters I'm during summer time in UTC+2, in winter in UTC+1.

### johndyer commented Apr 11, 2020

`getEaster(2020); `
returns
`[4, 12]`
for me...

### sunsero commented Apr 11, 2020

Thanks, you're right - if I pass 2020 as a constant value rather than in a variable it gives the correct result. I'll have to figure out what's going on.

### Termplexed commented Mar 1, 2021

Minor nuisance. On line 12 there is thin-space in stead of normal space. (`0x2009` instead of `0x20`) in `C - f`

`		H = (C<0x2009>-<0x2009>f(C / 4) - f((8 * C + 13)/25) + 19 * G + 15) % 30,`

### johndyer commented Mar 1, 2021

Thanks @Termplexed! Fixed.

### creat73 commented Apr 1, 2021

This algorithm seems to fail for year 2022 returning 23.04 instead of 17.04. Works fine for any other year I tried. Any idea on why that might be? I'm really curious and couldn't find explanation.

### wysal commented Jun 9, 2021 • edited Loading

I needed to count days up from easter for public holidays calculation so I did some adaptions:

function getEaster(year, plusdays) {
var f = Math.floor,
G = year % 19,
C = f(year / 100),
H = (C - f(C / 4) - f((8 * C + 13) / 25) + 19 * G + 15) % 30,
I = H - f(H / 28) * (1 - f(29 / (H + 1)) * f((21 - G) / 11)),
J = (year + f(year / 4) + I + 2 - C + f(C / 4)) % 7,
L = I - J,
month = 3 + f((L + 40) / 44),
day = L + 28 - 31 * f(month / 4);
return new Date(year, month - 1, day + plusdays);
}

so this would then output for example Easter Monday:
`getEaster(2022, 1)`

in case of need for someone

### tdiluzio commented Nov 22, 2021

Thanks @johndyer and @wysal 👍

### P06316 commented Nov 26, 2021 • edited Loading

This algorithm seems to fail for year 2022 returning 23.04 instead of 17.04. Works fine for any other year I tried. Any idea on why that might be? I'm really curious and couldn't find explanation.

I had the same issue and solved it by applying parseInt(myYear, 0) on the variable containing the year.

console.log(getEaster("2022")); //Returns [4, 23] => NOK
console.log(getEaster(2022)); //Returns [4, 17] => OK
console.log(getEaster(parseInt("2022", 0))); //Returns [4, 17] => OK

### PenkoMlakar commented Mar 18, 2022

Thanks @johndyer!

### TheDoomfire commented Nov 29, 2022

Thanks seem to work!

Was wondering if it's possible to get the good Friday from this? Since it's the Friday before Easter.

So I guess `day - 2` will do for most cases at least.

### CraigDavison commented Jan 20, 2023

When using this, I was passing the year in from an form input and it seems as though, only odd numbered years, returned the correct date (being Easter Sunday). The answer was to ensure the value passed in was an integer so added a line at the top of the function.

`year = parseInt(year)`

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