返回给定年份复活节日期的函数

Question

所以，这是一个有趣的小编程挑战。我正在编写一个快速方法来确定特定年份的所有市场假期，然后我开始阅读有关复活节的内容并发现确定其日期的逻辑是多么疯狂*——复活节满月后的第一个星期日春分之后！有人知道现有的函数可以计算给定年份的复活节日期吗？

当然，这可能并不那么难做到；我只是想如果有人已经这样做了我会问一下。（这似乎很有可能。）

更新

：实际上，我真的在寻找耶稣受难日（复活节之前之前的星期五）的日期......我只是认为复活节会让我到达那里。既然我在美国，我想我正在寻找天主教复活节？但如果我错了，也许有人可以纠正我。 *我所说的“疯狂”是指，

参与

。没有什么攻击性的...

Python：使用

Answer 1

easter()

函数。

>>> from dateutil.easter import *
>>> print easter(2010)
2010-04-04
>>> print easter(2011)
2011-04-24

函数获取您喜欢的计算类型作为参数：

EASTER_JULIAN   = 1
EASTER_ORTHODOX = 2
EASTER_WESTERN  = 3

您可以选择与美国相关的。

从结果中减少两天将为您带来耶稣受难日：

>>> from datetime import timedelta
>>> d = timedelta(days=-2)
>>> easter(2011)
datetime.date(2011, 4, 24)
>>> easter(2011)+d
datetime.date(2011, 4, 22)

奇怪的是，有人对此进行了迭代，并将结果发布在

维基百科关于该算法的文章中

：

alt text

在 SQL Server 复活节周日看起来像这样，向下滚动到耶稣受难日

Answer 2

CREATE FUNCTION dbo.GetEasterSunday 
( @Y INT ) 
RETURNS SMALLDATETIME 
AS 
BEGIN 
    DECLARE     @EpactCalc INT,  
        @PaschalDaysCalc INT, 
        @NumOfDaysToSunday INT, 
        @EasterMonth INT, 
        @EasterDay INT 

    SET @EpactCalc = (24 + 19 * (@Y % 19)) % 30 
    SET @PaschalDaysCalc = @EpactCalc - (@EpactCalc / 28) 
    SET @NumOfDaysToSunday = @PaschalDaysCalc - ( 
        (@Y + @Y / 4 + @PaschalDaysCalc - 13) % 7 
    ) 

    SET @EasterMonth = 3 + (@NumOfDaysToSunday + 40) / 44 

    SET @EasterDay = @NumOfDaysToSunday + 28 - ( 
        31 * (@EasterMonth / 4) 
    ) 

    RETURN 
    ( 
        SELECT CONVERT 
        (  SMALLDATETIME, 
                 RTRIM(@Y)  
            + RIGHT('0'+RTRIM(@EasterMonth), 2)  
            + RIGHT('0'+RTRIM(@EasterDay), 2)  
        ) 
    ) 

END 
GO

Good Friday

就是这样，它使用了上面的复活节功能

CREATE FUNCTION dbo.GetGoodFriday ( @Y INT ) RETURNS SMALLDATETIME AS BEGIN RETURN (SELECT dbo.GetEasterSunday(@Y) - 2) END GO

从这里：

http://web.archive.org/web/20070611150639/http://sqlserver2000.databases.aspfaq.com/why-should-i-consider-using-an-auxiliary-calendar-table.html

当我写这篇文章时（基于星期几和节假日的流量预测），我放弃了自己写的尝试。我在网上的某个地方找到了它。该代码是公共领域，但是...

Answer 3

叹息

你自己看看吧。

void dateOfEaster(struct tm* p)
{
    int Y = p->tm_year;
    int a = Y % 19;
    int b = Y / 100;
    int c = Y % 100;
    int d = b / 4;
    int e = b % 4;
    int f = (b + 8) / 25;
    int g = (b - f + 1) / 3;
    int h = (19 * a + b - d - g + 15) % 30;
    int i = c / 4;
    int k = c % 4;
    int L = (32 + 2 * e + 2 * i - h - k) % 7;
    int m = (a + 11 * h + 22 * L) / 451;
    p->tm_mon = ((h + L - 7 * m + 114) / 31 ) - 1;
    p->tm_mday = ((h + L - 7 * m + 114) % 31) + 1;
    p->tm_hour = 12;
    const time_t tmp = mktime(p);
    *p = *localtime(&tmp);  //recover yday from mon+mday
}

有些问题最好不要问。

我很幸运，我的国家所有的搬家假期都是从复活节日期开始的固定偏移量。

下面的 SQL Server 函数比接受的答案更通用

Answer 4

接受的答案仅适用于以下范围（含）：1900-04-15 至 2099-04-12

它使用美国海军天文台（USNO）提供的算法

http://aa.usno.navy.mil/faq/docs/easter.php

CREATE FUNCTION dbo.GetEasterSunday (@Y INT) RETURNS DATETIME AS BEGIN -- Source of algorithm : http://aa.usno.navy.mil/faq/docs/easter.php DECLARE @c INT = @Y / 100 DECLARE @n INT = @Y - 19 * (@Y / 19) DECLARE @k INT = (@c - 17) / 25 DECLARE @i INT = @c - @c / 4 - (@c - @k) / 3 + 19 * @n + 15 SET @i = @i - 30 * (@i / 30) SET @i = @i - (@i / 28) * (1 - (@i / 28) * (29 / (@i + 1)) * ((21 - @n) / 11)) DECLARE @j INT = @Y + @Y / 4 + @i + 2 - @c + @c / 4 SET @j = @j - 7 * (@j / 7) DECLARE @l INT = @i - @j DECLARE @m INT = 3 + (@l + 40) / 44 DECLARE @d INT = @l + 28 - 31 * (@m / 4) RETURN ( SELECT CONVERT ( DATETIME, RTRIM(@Y) + RIGHT('0'+RTRIM(@m), 2) + RIGHT('0'+RTRIM(@d), 2) ) ) END GO

希腊东正教和天主教复活节的 VB .NET 函数：

Answer 5

Public Shared Function OrthodoxEaster(ByVal Year As Integer) As Date
    Dim a = Year Mod 19
    Dim b = Year Mod 7
    Dim c = Year Mod 4
    Dim d = (19 * a + 16) Mod 30
    Dim e = (2 * c + 4 * b + 6 * d) Mod 7
    Dim f = (19 * a + 16) Mod 30
    Dim key = f + e + 3
    Dim month = If((key > 30), 5, 4)
    Dim day = If((key > 30), key - 30, key)
    Return New DateTime(Year, month, day)
End Function

Public Shared Function CatholicEaster(ByVal Year As Integer) As DateTime
    Dim month = 3
    Dim a = Year Mod 19 + 1
    Dim b = Year / 100 + 1
    Dim c = (3 * b) / 4 - 12
    Dim d = (8 * b + 5) / 25 - 5
    Dim e = (5 * Year) / 4 - c - 10
    Dim f = (11 * a + 20 + d - c) Mod 30
    If f = 24 Then f += 1
    If (f = 25) AndAlso (a > 11) Then f += 1
    Dim g = 44 - f
    If g < 21 Then g = g + 30
    Dim day = (g + 7) - ((e + g) Mod 7)
    If day > 31 Then
        day = day - 31
        month = 4
    End If
    Return New DateTime(Year, month, day)
End Function

以下代码通过powershell确定复活节：

Answer 6

function Get-DateOfEaster {
    param(
        [Parameter(ValueFromPipeline)]
        $theYear=(Get-Date).Year
        )

    if($theYear -lt 1583) {
        return $null
    } else {

        # Step 1: Divide the theYear by 19 and store the
        # remainder in variable A.  Example: If the theYear
        # is 2000, then A is initialized to 5.

        $a = $theYear % 19

        # Step 2: Divide the theYear by 100.  Store the integer
        # result in B and the remainder in C.

        $c = $theYear % 100
        $b = ($theYear -$c) / 100

        # Step 3: Divide B (calculated above).  Store the
        # integer result in D and the remainder in E.

        $e = $b % 4
        $d = ($b - $e) / 4

        # Step 4: Divide (b+8)/25 and store the integer
        # portion of the result in F.

        $f = [math]::floor(($b + 8) / 25)

        # Step 5: Divide (b-f+1)/3 and store the integer
        # portion of the result in G.

        $g = [math]::floor(($b - $f + 1) / 3)

        # Step 6: Divide (19a+b-d-g+15)/30 and store the
        # remainder of the result in H.

        $h = (19 * $a + $b - $d - $g + 15) % 30

        # Step 7: Divide C by 4.  Store the integer result
        # in I and the remainder in K.

        $k = $c % 4
        $i = ($c - $k) / 4

        # Step 8: Divide (32+2e+2i-h-k) by 7.  Store the
        # remainder of the result in L.

        $l = (32 + 2 * $e + 2 * $i - $h - $k) % 7

        # Step 9: Divide (a + 11h + 22l) by 451 and
        # store the integer portion of the result in M.

        $m = [math]::floor(($a + 11 * $h + 22 * $l) / 451)

        # Step 10: Divide (h + l - 7m + 114) by 31.  Store
        # the integer portion of the result in N and the
        # remainder in P.

        $p = ($h + $l - 7 * $m + 114) % 31
        $n = (($h + $l - 7 * $m + 114) - $p) / 31

        # At this point p+1 is the day on which Easter falls.
        # n is 3 for March and 4 for April.

        $DateTime = New-Object DateTime $theyear, $n, ($p+1), 0, 0, 0, ([DateTimeKind]::Utc)
        return $DateTime
    }
}

$eastersunday=Get-DateOfEaster 2015
Write-Host $eastersunday

在某处找到这个 Excel 公式

Answer 7

A1

包含年份，例如2020

ROUND(DATE(A1;4;1)/7+MOD(19*MOD(A1;19)-7;30)*0,14;0)*7-6

转换为 T-SQL 让我看到这个：

DECLARE @yr INT=2020
SELECT DATEADD(dd, ROUND(DATEDIFF(dd, '1899-12-30', DATEFROMPARTS(@yr, 4, 1)) / 7.0 + ((19.0 * (@yr % 19) - 7) % 30) * 0.14, 0) * 7.0 - 6, -2)

在 JS 中，取自

Answer 8

。

然后运行

getEaster(2020); // -> Sun Apr 12 2020

Oracle，给出了 1900 年至 2078 年之间的正确复活节日期：

Answer 9

to_date('18991230','yyyymmdd') + round((to_date(callendar_year||'04','yyyymm') -
to_date('18991230','yyyymmdd')) / 7 + mod(19 * MOD(callendar_year, 19) + 23, 30) * 0.14, 0) * 7-6

这可能会告诉您比您想知道的更多关于复活节日期是如何计算的信息。文末有BASIC中的函数

Answer 10

0
投票

https://www.assa.org.au/edm#Method

返回给定年份复活节日期的函数

问题描述投票：0回答：10

10个回答

最新问题

返回给定年份复活节日期的函数

问题描述 投票：0回答：10

10个回答

最新问题

问题描述投票：0回答：10