检查闰年

Question

如何检查一年是否是闰年？

我有这段代码：

declare @year int
set @year = 1968

SELECT CASE WHEN @YEAR = <LEAPYEAR> THEN 'LEAP YEAR' ELSE 'NORMAL YEAR' END

预期结果：

LEAP YEAR

Answer 1

检查2月29日：

CASE WHEN ISDATE(CAST(@YEAR AS char(4)) + '0229') = 1 THEN 'LEAP YEAR' ELSE 'NORMAL YEAR' END

或使用以下规则

CASE WHEN (@YEAR % 4 = 0 AND @YEAR % 100 <> 0) OR @YEAR % 400 = 0 THEN 'LEAP YEAR'...

Answer 2

闰年计算：

(@year % 4 = 0) and (@year % 100 != 0) or (@year % 400 = 0)

如果这是真的，那么这是闰年。或者将其放在案例陈述中

select case when
    (
        (@year % 4 = 0) and (@year % 100 != 0) or
        (@year % 400 = 0)
    ) then 'LEAP' else 'USUAL' end
;

Answer 3

最有效的一年中的测试：

CASE WHEN @YEAR & 3 = 0 AND (@YEAR % 25 <> 0 OR @YEAR & 15 = 0) THEN ...

改编自：http://stackoverflow.com/a/11595914/3466415

Answer 4

这也可以帮助

DECLARE @year INT = 2012

SELECT IIF(DAY(EOMONTH(DATEFROMPARTS(@year,2,1))) = 29,1,0)  
Result: 1 --(1 if Leap Year, 0 if not)

SELECT IIF(DAY(EOMONTH(DATEFROMPARTS(@year,2,1))) = 29,'Leap year','Not Leap year')  
Result: Leap year 

Answer 5

我有更好的解决方案

CREATE FUNCTION dbo.IsLeapYear(@year INT)
RETURNS BIT AS  
     BEGIN
          DECLARE @d DATETIME,
                  @ans BIT      
          SET @d = CONVERT(DATETIME,'31/01/'+CONVERT(VARCHAR(4),@year),103)
          IF DATEPART(DAY,DATEADD(MONTH,1,@d))=29 SET @ans=1 ELSE SET @ans=0
          RETURN @ans
    END
GO

随意使用

Answer 6

不确定将其与其他解决方案相比效率如何。但这是另一种选择。

DECLARE @year int = 2016
SELECT CASE 
  WHEN DATEPART(dayofyear, DATEFROMPARTS(@year, 12, 31)) = 366
  THEN 'LEAP' 
  ELSE 'NOT LEAP' 
END

Answer 7

3行...但也可能是2 ...

DECLARE @Y as int = 2021;
DECLARE @Dt as char(10) = CAST(@Y as CHAR(4)) + '-02-29'
SELECT IIF(isDATE(@Dt) = 1, 1,0)

或

DECLARE @Dt as char(10) = '2020-02-29'
SELECT IIF(isDATE(@Dt) = 1, 1,0)

艾伦

Answer 8

select decode(mod(&n,4),0,'leap year' ,'not a leapyear') as CHECK_LEAPYEAR  from dual

Answer 9

我疯了吗，但这不像

那样容易吗？

CASE WHEN (@YEAR % 4 = 0) THEN 'LEAP YEAR' ELSE 'NORMAL YEAR' END

Answer 10

select

CASE

  WHEN result = 0 THEN 'Leap_Year'

  WHEN result <> 0 THEN 'Not_A_Leap_Year'

END

from(select mod((EXTRACT(YEAR FROM DATE '2013-08-23')), 4) result FROM DUAL);