0

我在这个网站上得到了一年中所有星期的代码，如下所示，我应该填充星期日期，开始日期为星期六，结束日期为星期五。当一周结束时，它应该进入下一周的日期。我怎么能做到这一点，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) -- 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];
4

2 回答 2

0

不太确定你在问什么，但根据你上面的查询，这将给出周数，基于星期六是一周的第一天，2013 年：

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，根据各种评论和答案，我认为这里需要的是，对于给定的一天，返回同一周的所有天，星期六作为一周的第一天。所以是这样的：

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

这给出了结果：

在此处输入图像描述

我认为这是这里需要的吗？

第二次编辑：

首先，创建函数：

create function dbo.weekDates (@date date)
returns @dates table ([date] date, [dayofweek] varchar(9))
as
begin

  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)
  )
  insert into @dates ([date], [dayofweek])
  select [date], [dayOfWeek] = datename(dw, [date])
  from daysOfWeek;

  return

end

go

使用功能：

set datefirst 6 -- Set Saturday as first day of week
select * from dbo.weekDates (getdate()) -- Change input parameter as required
于 2013-02-13T14:24:09.190 回答
0

我已经实现了显示周日期，但它正在显示下周日期我想显示当前周日期，这里开始日是星期六，结束日是星期五

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
于 2013-02-14T06:50:26.570 回答