在sql中按当前日期查找星期的日期范围？

Question

我想在本周找到记录,只是宣布当前日期.例:

     select datename(dw,getdate())  -- Example today is Friday "27-04-2012"

那么我怎样才能获得日期范围

     start on monday "23-04-2012" or sunday "22-04-2012"  As @dateStart to

     end on sunday "29-04-2012"  or saturday "28-04-2012" As @dateEnd

然后我可以选择查询

     select * from table where date>=@dateStart  AND date<=@dateEnd

Answer 1

这里有一些有用的命令来获取一周中的每一天

SELECT 
    DATEADD(wk, DATEDIFF(wk,0,GETDATE()), -2) SatOfPreviousWeek,
    DATEADD(wk, DATEDIFF(wk,0,GETDATE()), -1) SunOfCurrentWeek,
    DATEADD(wk, DATEDIFF(wk,0,GETDATE()), 0) MondayOfCurrentWeek,
    DATEADD(wk, DATEDIFF(wk,0,GETDATE()), 1) TuesOfCurrentWeek,
    DATEADD(wk, DATEDIFF(wk,0,GETDATE()), 2) WedOfCurrentWeek,
    DATEADD(wk, DATEDIFF(wk,0,GETDATE()), 3) ThursOfCurrentWeek,
    DATEADD(wk, DATEDIFF(wk,0,GETDATE()), 4) FriOfCurrentWeek,
    DATEADD(wk, DATEDIFF(wk,0,GETDATE()), 5) SatOfCurrentWeek

然后,您将在查询中使用这些来获取日期范围:

SELECT *
FROM yourTable
WHERE yourDate >= DATEADD(wk, DATEDIFF(wk,0,GETDATE()), -1) -- Sunday
AND yourDate <= DATEADD(wk, DATEDIFF(wk,0,GETDATE()), 5) -- Saturday

Answer 2

所以你只想要本周的记录？

SELECT t.* 
FROM table t 
WHERE t.date >= DATEADD(DAY, DATEDIFF(DAY, 0, GETDATE()) / 7 * 7, 0)
AND   t.date <= DATEADD(DAY, DATEDIFF(DAY, -1, GETDATE()), 0)