Sql server 2008 将整个星期的开始日期显示为星期六
我在这个网站上得到了一年中所有周的代码,如下所示,我应该填充周日期,开始日期为星期六,结束日期为星期五。当一周结束时,应进入下一周,并注明日期。 我怎样才能做到这一点请帮助我Sql server 2008 将整个星期的开始日期显示为星期六,sql-server-2008,Sql Server 2008,我在这个网站上得到了一年中所有周的代码,如下所示,我应该填充周日期,开始日期为星期六,结束日期为星期五。当一周结束时,应进入下一周,并注明日期。 我怎样才能做到这一点请帮助我 DECLARE @Year INT=2013; DECLARE @start DATE; --DECLARE @WK INT=2 SET @start = DATEADD(YEAR, @Year-1900, 0); ;WITH n AS ( SELECT TOP (366
DECLARE @Year INT=2013;
DECLARE @start DATE;
--DECLARE @WK INT=2
SET @start = DATEADD(YEAR, @Year-1900, 0);
;WITH n AS
(
SELECT TOP (366) -- in case of leap year
TDate = DATEADD(DAY, ROW_NUMBER() OVER (ORDER BY name)-1, @start)
FROM sys.all_objects
),
x AS
(
SELECT md = MIN(TDate) FROM n
WHERE DATEPART(WEEKDAY, TDate) = 7 -- assuming DATEFIRST is SATURDAY
),
y(TDate,wk) AS
(
SELECT n.TDate, ((DATEPART(DAYOFYEAR,n.TDate)-
DATEDIFF(DAY, @start,x.md)-1)/7)+1
FROM n CROSS JOIN x
WHERE n.TDate >= x.md
AND n.TDate < DATEADD(YEAR, 1, @start)
)
SELECT [date] = TDate, [week] = wk
FROM y WHERE wk < 53
ORDER BY [date];
我不太清楚你在问什么,但根据你上面的查询,这将给出2013年的周数,以星期六为一周的第一天:
DECLARE @Year INT=2013;
DECLARE @start DATE;
SET @start = DATEADD(YEAR, @Year-1900, 0);
SET DATEFIRST 6; -- Set start of week as Saturday
WITH n AS
(
SELECT TOP (366) -- in case of leap year
TDate = DATEADD(DAY, ROW_NUMBER() OVER (ORDER BY name)-1, @start)
FROM sys.all_objects
)
select TDate
, DATEPART(WEEK,TDate)
from n
where year(TDate) = 2013;
编辑:
因此,根据各种评论和回答,我认为这里需要的是,对于给定的一天,返回同一周中的所有天,周六为一周的第一天。比如说:
set datefirst 6; -- make sure first day of week is Saturday
declare @date date = getdate(); -- change date as required here
with daysOfWeek as
(
select [date] = dateadd(dd, -datepart(dw, @date) + 1, @date)
union all
select [date] = dateadd(dd, 1, [date])
from daysOfWeek
where [date] < dateadd(dd, -datepart(dw, @date) + 7, @date)
)
select [date], dayOfWeek = datename(dw, [date])
from daysOfWeek
我已经实现了显示一周的日期,但它正在显示下周的日期。我想显示当前一周的日期,这里的开始日期是星期六,结束日期是星期五
ALTER FUNCTION GetCurrentWeek()
RETURNS @TWeek TABLE (TWeek NVARCHAR(20))
AS
BEGIN
DECLARE @Year INT= DATEPART(YEAR,GETDATE());
DECLARE @start DATE;
SET @start = DATEADD(YEAR, @Year-1900, 0);
;WITH n AS
(
SELECT TOP (366) -- in case of leap year
TDate = DATEADD(DAY, ROW_NUMBER() OVER (ORDER BY name)-1, @start)
FROM sys.all_objects
),
x AS
(
SELECT md = MIN(TDate) FROM n
WHERE DATEPART(WEEKDAY, TDate) = 7 -- assuming DATEFIRST is SATURDAY
),
y(TDate,wk) AS
(
SELECT n.TDate, ((DATEPART(DAYOFYEAR, n.TDate)
- DATEDIFF(DAY, @start, x.md)-1)/7) + 1
FROM n CROSS JOIN x
WHERE n.TDate >= x.md
AND n.TDate < DATEADD(YEAR, 1, @start)
)
INSERT @TWeek
SELECT [date] = TDate
FROM y WHERE wk =DATEPART(wk, GetDate())
ORDER BY [date];
RETURN;
END
我不明白你的问题,你希望得到什么样的产出?您是否考虑过使用复杂查询而不是复杂查询?或者这是为了填充日历表?我已经实现了显示周日期,但它显示了我希望显示当前周日期的下周日期。请参见上面的编辑。如果这仍然不正确,您确实需要发布特定输入所需的输出
set datefirst 6 -- Set Saturday as first day of week
select * from dbo.weekDates (getdate()) -- Change input parameter as required
ALTER FUNCTION GetCurrentWeek()
RETURNS @TWeek TABLE (TWeek NVARCHAR(20))
AS
BEGIN
DECLARE @Year INT= DATEPART(YEAR,GETDATE());
DECLARE @start DATE;
SET @start = DATEADD(YEAR, @Year-1900, 0);
;WITH n AS
(
SELECT TOP (366) -- in case of leap year
TDate = DATEADD(DAY, ROW_NUMBER() OVER (ORDER BY name)-1, @start)
FROM sys.all_objects
),
x AS
(
SELECT md = MIN(TDate) FROM n
WHERE DATEPART(WEEKDAY, TDate) = 7 -- assuming DATEFIRST is SATURDAY
),
y(TDate,wk) AS
(
SELECT n.TDate, ((DATEPART(DAYOFYEAR, n.TDate)
- DATEDIFF(DAY, @start, x.md)-1)/7) + 1
FROM n CROSS JOIN x
WHERE n.TDate >= x.md
AND n.TDate < DATEADD(YEAR, 1, @start)
)
INSERT @TWeek
SELECT [date] = TDate
FROM y WHERE wk =DATEPART(wk, GetDate())
ORDER BY [date];
RETURN;
END