Question

除了1月1日，SQL Server显示2011年第一周的第53周，需要是第1周。

以下是查询和输出：

declare @T table (dt datetime)
insert into @T values
('2010-12-26'),
('2010-12-27'),
('2010-12-28'),
('2010-12-29'),
('2010-12-30'),
('2010-12-31'),
('2011-01-01'),
('2011-01-02'),
('2011-01-03'),
('2011-01-04'),
('2011-01-05'),
('2011-01-06'),
('2011-01-07'),
('2011-01-08')
select dt,DATEPART(wk,dt) from @T

输出：

2010-12-26 00:00:00.000 53
2010-12-27 00:00:00.000 53
2010-12-28 00:00:00.000 53
2010-12-29 00:00:00.000 53
2010-12-30 00:00:00.000 53
2010-12-31 00:00:00.000 53
2011-01-01 00:00:00.000 1
2011-01-02 00:00:00.000 2
2011-01-03 00:00:00.000 2
2011-01-04 00:00:00.000 2
2011-01-05 00:00:00.000 2
2011-01-06 00:00:00.000 2
2011-01-07 00:00:00.000 2
2011-01-08 00:00:00.000 2

我希望SQL Server在12月26日 - 1月1日显示第1周。有人知道如何做到这一点吗？

谢谢和问候， Aschwin。

Answer 1

这比我最初预期的要难得多。我比较去年年底，看看它是否有资格参与新的一年。如果是这样，我将周设置为1，否则我只使用正常的一周。

declare @T table (dt datetime) 
insert into @T values 
('2010-12-25'), 
('2010-12-26'), 
('2010-12-27'), 
('2010-12-28'), 
('2010-12-29'), 
('2010-12-30'), 
('2010-12-31'), 
('2011-01-01'), 
('2011-01-02'), 
('2011-01-03'), 
('2011-01-04'), 
('2011-01-05'), 
('2011-01-06'), 
('2011-01-07'), 
('2011-01-08'), 
('2011-12-31'),
('2012-01-01')

select dt, 
week = case when dt + 6 - datediff(day, -1, dt) % 7 = dateadd(year, datediff(year,-1, dt), 0)
then 1 else datepart(week, dt) end from @t

证明：

http://data.stackexchange.com/stackoverflow/q/110527/

Answer 2

我不确定它是否适用于所有年份（但它看起来像）但您可以使用CASE语句解决此问题。

SELECT  dt
        , CASE  WHEN DATEPART(wk, dt) <> 53 
                THEN DATEPART(wk, dt) 
                ELSE 1 
          END  
FROM    @T

新的ISO_WEEK日期部分不适用于您请求的输出。

Answer 3

我创建了2个函数来处理这个问题 1）获得一周的第一天或最后一天 2）获得周数或年份

功能1

    CREATE FUNCTION [dbo].[fn_GetDayOf]
(
    @Date datetime,
    --@FirstDayOfWeek int = 7,
    @Mode int =1
)
/*
    Mode 1: First Day Of Week
    Mode 2: Last Day Of Week
*/

RETURNS datetime
WITH EXECUTE AS CALLER
BEGIN
    Declare @Return datetime

    --SET DATEFIRST @FirstDayOfWeek

    IF @Mode = 1
    BEGIN
        select @Return = dateadd(day,-(datepart(weekday,@date)-1),convert(date,@date))
    END
    ELSE IF @Mode = 2
    BEGIN
        select @Return = dateadd(SECOND,-1,convert(datetime,dateadd(day,(datepart(weekday,@date)),convert(date,@date))))
    END
    ELSE
    BEGIN
        SET @Return = @Date
    END
    --SET DATEFIRST 7

    RETURN @Return
END

功能2

    CREATE FUNCTION [dbo].[fn_GetYearWeek]
(
    @Date datetime,
    --@FirstDayOfWeek int = 7,
    @Mode int =1
)
/*
    Mode 1 = Week Number
    Mode 2 = Year
*/
RETURNS INT
BEGIN
    declare @Return int
    IF @Mode = 1
    BEGIN
        select @Return = case when  datepart(week,[dbo].[fn_GetDayOf] (@Date,1)) <> datepart(week,[dbo].[fn_GetDayOf] (@Date,2)) then datepart(week,[dbo].[fn_GetDayOf] (@Date,1)) else datepart(week,[dbo].[fn_GetDayOf] (@Date,2)) end 
    END
    ELSE IF @Mode = 2
    BEGIN
        select @Return = case when  datepart(WEEK,[dbo].[fn_GetDayOf] (@Date,1)) <> datepart(week,[dbo].[fn_GetDayOf] (@Date,2)) then datepart(YEAR,[dbo].[fn_GetDayOf] (@Date,1)) else datepart(YEAR,[dbo].[fn_GetDayOf] (@Date,2)) end 
    END
    ELSE
    BEGIN
        SET @Return = -1
    END

    Return  @Return
END

运行示例

declare @T table (dt datetime) 
insert into @T values 
('2010-12-25'), 
('2010-12-26'), 
('2010-12-27'), 
('2010-12-28'), 
('2010-12-29'), 
('2010-12-30'), 
('2010-12-31'), 
('2011-01-01'), 
('2011-01-02'), 
('2011-01-03'), 
('2011-01-04'), 
('2011-01-05'), 
('2011-01-06'), 
('2011-01-07'), 
('2011-01-08'), 
('2011-12-31'),
('2012-01-01'),
('2012-01-02'),
('2012-12-31'),
('2013-01-01')

select 
    dt,
    datepart(week,dt),
    --case when  datepart(week,[dbo].[fn_GetDayOf] (dt,1)) <> datepart(week,[dbo].[fn_GetDayOf] (dt,2)) then datepart(week,[dbo].[fn_GetDayOf] (dt,1)) else datepart(week,[dbo].[fn_GetDayOf] (dt,2)) end 
    [dbo].[fn_GetYearWeek] (dt,1),
    [dbo].[fn_GetYearWeek] (dt,2)
from @T

结果：

Answer 4

检索当年总周数的另一种方法：

DECLARE @LASTDAY DATETIME
DECLARE @weeks INT
SET @LASTDAY = DATEADD(ms,-3,DATEADD(yy,0,DATEADD(yy,DATEDIFF(yy,0,GETDATE())+1,0)))
SELECT @weeks = CASE DATEname(dw,@LASTDAY)

    WHEN 'MONDAY' THEN DATEPART(WK, DATEADD(wk,DATEDIFF(wk,7,@LASTDAY),5))
    WHEN 'TUESDAY' THEN DATEPART(WK, DATEADD(wk,DATEDIFF(wk,7,@LASTDAY),5))
    WHEN 'WEDNESDAY' THEN DATEPART(WK, DATEADD(wk,DATEDIFF(wk,7,@LASTDAY),5))
    ELSE DATEPART(WK, @LASTDAY)
END
select @weeks

SQL Server未显示给定日期的正确周数

4 个答案: