easter_date

(PHP 4, PHP 5)

easter_date Retorna a data do sistema UNIX correspondente à meia-noite da Páscoa do ano especificado. Se nenhum ano tiver sido especificado, será assumido o ano atual.

Descrição

int easter_date ([ int $ano ] )

Retorna a data do sistema UNIX correspondente à meia-noite da Páscoa do ano especificado. Se nenhum ano tiver sido especificado, será assumido o ano atual.

À partir da versão 4.3.0 PHP, e se omitido o parâmetro ano, o padrão assumido é o ano atual de acordo com a hora local.

Aviso Esta função irá gerar um aviso se o ano está fora do padrão de datas do sistema UNIX (i.e. antes de 1970 ou após 2037).

Exemplo #1 easter_date() exemplo

<?php

echo date ("M-d-Y"easter_date(1999));        // Apr-04-1999
echo date ("M-d-Y"easter_date(2000));        // Apr-23-2000
echo date ("M-d-Y"easter_date(2001));        // Apr-15-2001

?>

A data da Páscoa foi definida pelo Conselho de Nicaea em DC325 como sendo o primeiro domingo após a primeira lua cheia que cai no equinócio da primavera ou depois dele. O equinócio geralmente cai perto do 21º de Março, logo, o cálculo resume-se à determinar a data da lua cheia e a data do seguinte domingo. O algoritmo usado aqui foi feito no ano 532 por Dionysius Exiguus. De acordo com o calendário "Julian" (para anos anterioris à 1753) um simples ciclo de 19-anos é usado para seguir as fases da lua. De acordo com o Calendário Gregoriano (para antes após 1753 - planejado por Clavius e por Lilius, e feito por Pope Gregory XIII em Outubro de 1582, e na Grã Bretanha e suas colônias em Setembro de 1752), duas correções fatoriais foram adicionadas para fazer o ciclo mais exato.

(O código é baseado em um programa feito em C por Simon Kershaw, <webmaster@ely.anglican.org>)

Veja easter_days() para calcular a Páscoa antes de 1970 ou após 2037.

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User Contributed Notes 9 notes

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3
py dot lebecq at gmail dot com
4 years ago
I recently had to write a function that allows me to know if today is a holiday.

And in France, we have some holidays which depends on the easter date. Maybe this will be helpful to someone.

Just modify in the $holidays array the actual holidays dates of your country.

<?php
/**
* This function returns an array of timestamp corresponding to french holidays
*/
protected static function getHolidays($year = null)
{
  if (
$year === null)
  {
   
$year = intval(date('Y'));
  }
   
 
$easterDate  = easter_date($year);
 
$easterDay   = date('j', $easterDate);
 
$easterMonth = date('n', $easterDate);
 
$easterYear   = date('Y', $easterDate);

 
$holidays = array(
   
// These days have a fixed date
   
mktime(0, 0, 0, 11$year),  // 1er janvier
   
mktime(0, 0, 0, 51$year),  // Fête du travail
   
mktime(0, 0, 0, 58$year),  // Victoire des alliés
   
mktime(0, 0, 0, 714, $year),  // Fête nationale
   
mktime(0, 0, 0, 815, $year),  // Assomption
   
mktime(0, 0, 0, 11, 1$year),  // Toussaint
   
mktime(0, 0, 0, 11, 11, $year),  // Armistice
   
mktime(0, 0, 0, 12, 25, $year),  // Noel

    // These days have a date depending on easter
   
mktime(0, 0, 0, $easterMonth, $easterDay + 2$easterYear),
   
mktime(0, 0, 0, $easterMonth, $easterDay + 40, $easterYear),
   
mktime(0, 0, 0, $easterMonth, $easterDay + 50, $easterYear),
  );

 
sort($holidays);
 
  return
$holidays;
}
?>
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2
maxie
6 years ago
To compute the correct Easter date for Eastern Orthodox Churches I made a function based on the Meeus Julian algorithm:

<?php
function orthodox_eastern($year) {
   
$a = $year % 4;
   
$b = $year % 7;
   
$c = $year % 19;
   
$d = (19 * $c + 15) % 30;
   
$e = (2 * $a + 4 * $b - $d + 34) % 7;
   
$month = floor(($d + $e + 114) / 31);
   
$day = (($d + $e + 114) % 31) + 1;
   
   
$de = mktime(0, 0, 0, $month, $day + 13, $year);
   
    return
$de;
}
?>
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1
Guillaume Dufrene
1 year ago
I found a problem with holidays timestamp computation and daylight saving time.
An article about it at http://goo.gl/76t31 (in french only, sorry).

In summary, this year (2013) easter begins before adding an hour for daylight saving time (occured sunday at 3:00). It means that if you do $easter + X, where x is a number of seconds equivalent to one day, 39 days or 50 days, the result is not equals to a midnight timestamp...

Here a function to check if a midnight timestamp is equals to an holiday :

function isHoliday( $ts ) {
// Licence : Creative Commons (BY)
// By Webpulser - http://goo.gl/76t31
  $fixed_holidays = array( ’01-01′, ’01-05′, ’08-05′, ’14-07′, ’15-08′, ’11-11′, ’25-12′ );
  $format = ‘d-m’;

$dm = date($format, $ts);
  if ( in_array($dm, $fixed_holidays) ) return true;

$easter = easter_date( date(‘Y’, $ts) );
  if ( date($format, $easter +   86400) == $dm ) return true;
  if ( date($format, $easter + 3369600) == $dm ) return true;
  if ( date($format, $easter + 4320000) == $dm ) return true;

return false;
}

feel free to use / modify.
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0
adwil at live dot com
3 months ago
Hey, recently I needed a function to get realization dates in online shop, so here it is (ready to go for polish users, please adjust your dates for any other country):

<?php
function getWorkday($date1,$workDays) {
       
$workDays = (int)$workDays;
        if (
$workDays <= 0)
            return
null;

       
$date1=strtotime('-1 day',strtotime($date1));

       
$lastYear = null;
       
$hol=array('01-01','01-06','05-01','05-03','08-15','11-01','11-11','12-25','12-26');  //array of month-date of static holidays (these are from Poland)
       
$i = 0;
        while (
$i<=$workDays) {
           
$year = date('Y', $date1);
            if (
$year !== $lastYear){
               
$lastYear = $year;
               
$easter = date('m-d', easter_date($year));
               
$date = strtotime($year . '-' . $easter); // easter
               
$easterSec = date('m-d', strtotime('+1 day', $date)); // easter monday
               
$greens = date('m-d', strtotime('+49 days', $date)); // zielone swiatki
               
$cc = date('m-d', strtotime('+60 days', $date)); // boze cialo
               
$hol[] = $easter;
               
$hol[] = $easterSec;
               
$hol[] = $greens;
               
$hol[] = $cc;
            }
           
$weekDay=date('w',$date1);
            if (!(
$weekDay==0 || $weekDay==6 || in_array(date('m-d',$date1),$hol)))
               
$i++;

           
$date1=strtotime('+1 day',$date1);
        }
        return
date('Y-m-d',$date1);
    }
?>
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0
peter dot burden at gmail dot com
4 years ago
In 5.3.1 easter_date() returns GMT of start of Easter Day in UK allowing
for UK Summer Time. If you are in another time zone you need to
calculate offsets.
<?php
        $e1
= easter_date(2008);
       
$e2 = easter_date(2009);
        echo
"Timestamps " . $e1 . " " . $e2 . "\n";

       
//      Timestamps 1206230400 1239490800

       
echo "Days between Easter 2008 Easter 2009 = " . ($e2-$e1)/86400 . "\n";

       
//      Days between Easter 2008 Easter 2009 = 384.958333333 - i.e. 384 days 23 hours

       
date_default_timezone_set("Europe/London");
        echo
date( " l, jS F Y H:i TO",$e1) . "\n";
        echo
date( " l, jS F Y H:i TO",$e2) . "\n";

       
//      Sunday, 23rd March 2008 00:00 GMT+0000
        //      Sunday, 12th April 2009 00:00 BST+0100

       
date_default_timezone_set("America/New_York");
        echo
date( " l, jS F Y H:i TO",$e1) . "\n";
        echo
date( " l, jS F Y H:i TO",$e2) . "\n";

       
//      Saturday, 22nd March 2008 20:00 EDT-0400
        //      Saturday, 11th April 2009 19:00 EDT-0400
        //      Daylight saving time in effect - New York 4 hours behind GMT
?>
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0
nigelf at esp dot co dot uk
6 years ago
v5.2.1 - There is a known bug with easter_date() that can return incorrect dates for some years:

<?php
// 2008 OK
echo date("D d M Y", easter_date(2008));  // Sun 23 Mar 2008

// 2009 returns Saturday
echo date("D d M Y", easter_date(2009));  // Sat 11 Apr 2009
?>


However easter_days() works correctly:

<?php
echo date("D d M Y", strtotime("2009-03-21 +".easter_days(2009)." days"));  // Sun 12 Apr 2009
?>

It is apparently related to timezone settings.
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0
Anonymous
7 years ago
I made the function like this ... works fine !

<?php
function ostern
{
   
$maerz21=date('z',mktime(0,0,0,3,21,$jb));

   
$d=((15 + $jb/100 - $jb/400 - (8 * $jb/100 + 13) / 25)%30 + 19 * ($jb%19))%30;

    if (
$d==29)
    {
       
$D=28;
    }
    elseif(
$d==28 && ($jb%17)>=11)
    {
       
$D=27;
    }
    else
$D=$d;   

   
$e=(2 * ($jb%4) + 4 *($jb%7) + 6 * $D + (6 + $jb/100 - $jb/400 - 2)%7)%7;

   
$ostersonntag=$e+$D+1+$maerz21;
    return
$ostersonntag;
}
?>
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0
phpuser
9 years ago
The algorithm from Bigtree is correct if you add some (int) cast
<?php
  
function easter_date ($Year) {
  
      
/*
       G is the Golden Number-1
       H is 23-Epact (modulo 30)
       I is the number of days from 21 March to the Paschal full moon
       J is the weekday for the Paschal full moon (0=Sunday,
         1=Monday, etc.)
       L is the number of days from 21 March to the Sunday on or before
         the Paschal full moon (a number between -6 and 28)
       */
      

        
$G = $Year % 19;
        
$C = (int)($Year / 100);
        
$H = (int)($C - (int)($C / 4) - (int)((8*$C+13) / 25) + 19*$G + 15) % 30;
        
$I = (int)$H - (int)($H / 28)*(1 - (int)($H / 28)*(int)(29 / ($H + 1))*((int)(21 - $G) / 11));
        
$J = ($Year + (int)($Year/4) + $I + 2 - $C + (int)($C/4)) % 7;
        
$L = $I - $J;
        
$m = 3 + (int)(($L + 40) / 44);
        
$d = $L + 28 - 31 * ((int)($m / 4));
        
$y = $Year;
        
$E = mktime(0,0,0, $m, $d, $y);

         return
$E;

   }
?>
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-1
luca at musician dot org
2 years ago
Due to range limitations of core functions such as easter_date(), mktime(), strtotime(), I wrote this function to get the Easter Day in string format yyyy-mm-dd.

function myEaster($year)
{
    /*Warning: easter_date(): This function is only valid for years between 1970 and 2037
     * The easter_days() function can be used instead of easter_date() to calculate Easter for years which fall outside the range.
      */
     //The next line would do the work if there were no limitations:
    //return date("Y-m-d",easter_date($year));

    /*Outside range (1970,2037) they advise to use easter_days().
     * Unfortunately, when you have to create a date object as 21-03-yyyy to which add easter_days(), then obtain Easter,
     * functions like strtotime(), DateTime::createFromFormat() will fail. (return value is 01-01-1970)
     */
    $march21=date("$year-03-21");
    $days=easter_days($year);
    if($year<=2037)
    //The next line would do the work if strtotime() wasn't affected by same limitations. But, the if..else is required to handle all years.
        $date = date("Y-m-d",strtotime(date("Y-m-d", strtotime($march21)) . " +$days day"));
    else
    {
        if($days<=10){
            $day=str_pad(21+$days, 2, '0', STR_PAD_LEFT);
            $date=date("$year-03-$day");
        }
        else
        {
            $day=str_pad($days-10, 2, '0', STR_PAD_LEFT);
            $date=date("$year-04-$day");
        }
    }

    return $date;
}
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