显示整个星期的开始日期为星期六

我已经在一年中的所有星期在这个网站的代码如下，我应该填充星期日期，开始日期为星期六和结束日期为星期五。当一周结束时，它应该进入下一周的日期。我怎么能达到这个请帮助我。显示整个星期的开始日期为星期六

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];

来源

2013-02-13 john

我不明白你的问题，你期望输出什么？你有没有考虑过使用[日历表]（http://sqlserver2000.databases.aspfaq.com/why-should-i-consider-using-an-auxiliary-calendar-table.html）而不是复杂的查询？或者这是为了填充日历表？ – Pondlife 2013-02-13 17:14:27

不是真的知道你是问什么，但根据您的查询上面这将给本周的数字，根据上周六是一周的第一天，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，基于各种评论和答案我认为什么这里需要，对于给定的一天，返回所有天同一周，周六是一周的第一天。因此，像这样：

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

其中给出的结果：

enter image description here

我觉得这是怎么在这里需要？

第二个编辑：

首先，创建功能：

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

来源

2013-02-13 14:24:09

我已经实现了展示周日期，但是它显示了下周的日期，我想展示当前的周日期。 – john 2013-02-14 06:45:28

请参阅上述编辑。如果这仍然不正确，你*真的需要发布你需要输出的特定输出。 – 2013-02-14 10:52:38

最后的代码很有帮助，但是如何通过功能来实现 – john 2013-02-15 07:17:36

我已经实现了，以显示星期日期，但它显示下周我希望展示当前周日期，这里的开始日是星期六和结束日星期五

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

来源

2013-02-14 06:50:26 john

显示整个星期的开始日期为星期六

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