如何在sql中获取上周的最后一天？

答

无论实际DATEFIRST设置的，最后一个周日可以发现这样的：

SELECT DATEADD(day, 
       -1 - (DATEPART(weekday, GETDATE()) + @@DATEFIRST - 2) % 7, 
       GETDATE() 
      ) AS LastSunday

更换GETDATE()用参数@date获得特定日期之前的最后一个星期日。

+0

哇，上周我想出了在给定日期后的下一个星期五找到模数技巧时，我觉得我很聪明。先生，我向你致敬。对于非数学爱好者来说，Andriy正在使用余数来返回星期天以来的天数。我给了一个大拇指！ – 2011-05-09 21:21:56

答

SET @EndDate = GETDATE()-DatePart(dw, GETDATE());

+3

这不会返回正确的日期。现在运行它将返回上周六的4/30。 – IAmTimCorey 2011-05-08 00:04:16

答

这里是如何做到这一点有很大的文章：

http://www.objectreference.net/post/SQL-Find-last-week-date-range.aspx

你想使用@StartOfPrevWeek变量。

答

上周日（这是 “最后一周”结束）

SELECT DATEADD(wk, DATEDIFF(wk, 6, CURRENT_TIMESTAMP), 6) AS LAST_SUNDAY

本周（假设周一至周日周格式）

SELECT DATEADD(wk, DATEDIFF(wk, 7, CURRENT_TIMESTAMP), 7) AS START_OF_WEEK 
SELECT DATEADD(wk, DATEDIFF(wk, 6, CURRENT_TIMESTAMP), 6 + 7) AS END_OF_WEEK

结果

倍

START_OF_WEEK 
----------------------- 
2011-05-02 00:00:00.000 

END_OF_WEEK 
----------------------- 
2011-05-08 00:00:00.000

的例子来解释巫术（使用这上面的SQL更改为需要从本周开始和周结束日的一周）一周的下方

例子找到天在本周内（周日至周六）
如果实际的END_OF_WEEK是下一个周日到周六，那么您需要+7到本周的值。（见END_OF_WEEK上面的例子）

SQL下面

SELECT DATEADD(wk, DATEDIFF(wk, -2, CURRENT_TIMESTAMP), -2) AS DAY_OF_WEEK /* Saturday */ 
SELECT DATEADD(wk, DATEDIFF(wk, -1, CURRENT_TIMESTAMP), -1) AS DAY_OF_WEEK /* Sunday */ 
SELECT DATEADD(wk, DATEDIFF(wk, 0, CURRENT_TIMESTAMP), 0) AS DAY_OF_WEEK /* Monday */ 
SELECT DATEADD(wk, DATEDIFF(wk, 1, CURRENT_TIMESTAMP), 1) AS DAY_OF_WEEK /* Tuesday */ 
SELECT DATEADD(wk, DATEDIFF(wk, 2, CURRENT_TIMESTAMP), 2) AS DAY_OF_WEEK /* Wednesday */ 
SELECT DATEADD(wk, DATEDIFF(wk, 3, CURRENT_TIMESTAMP), 3) AS DAY_OF_WEEK /* Thursday */ 
SELECT DATEADD(wk, DATEDIFF(wk, 4, CURRENT_TIMESTAMP), 4) AS DAY_OF_WEEK /* Friday */ 
SELECT DATEADD(wk, DATEDIFF(wk, 5, CURRENT_TIMESTAMP), 5) AS DAY_OF_WEEK /* Saturday */ 
SELECT DATEADD(wk, DATEDIFF(wk, 6, CURRENT_TIMESTAMP), 6) AS DAY_OF_WEEK /* Sunday */ 
SELECT DATEADD(wk, DATEDIFF(wk, 7, CURRENT_TIMESTAMP), 7) AS DAY_OF_WEEK /* Monday */ 
SELECT DATEADD(wk, DATEDIFF(wk, 8, CURRENT_TIMESTAMP), 8) AS DAY_OF_WEEK /* Tuesday */ 
SELECT DATEADD(wk, DATEDIFF(wk, 9, CURRENT_TIMESTAMP), 9) AS DAY_OF_WEEK /* Wednesday */ 
SELECT DATEADD(wk, DATEDIFF(wk, 10, CURRENT_TIMESTAMP), 10) AS DAY_OF_WEEK /* Thursday */ 
SELECT DATEADD(wk, DATEDIFF(wk, 11, CURRENT_TIMESTAMP), 11) AS DAY_OF_WEEK /* Friday */ 
SELECT DATEADD(wk, DATEDIFF(wk, 12, CURRENT_TIMESTAMP), 12) AS DAY_OF_WEEK /* Saturday */ 
etc...

答

对于合适的日历表，SQL更直接。没有巫术。

select max(cal_date) end_of_last_week 
from calendar 
where (cal_date < current_date and day_of_week = 'Sun'); 

end_of_last_week 
-- 
2011-05-01

答

SELECT (DATEADD(DAY, ((DATEPART(dw, @Date) - 1) * -1), @Date))

答

DECLARE @LastSunday DATETIME 

-- This will get the previous Sunday with time as 23:59:59 
SELECT @LastSunday = Dateadd(SECOND, -1, Dateadd(WK, Datediff(WK, 6, 
                CURRENT_TIMESTAMP) 
               , 7)) 

SELECT @LastSunday 

-- This gets the monday prior to it and time of 00:00:00 
SELECT Dateadd(SECOND, 1, Dateadd(DAY, -7, @LastSunday)) 
-- This will make you time spans between eg, Monday 16/07/2012 00:00:00 through to Sunday 22/07/2012 23:59:59 
-- Then use them in your WHERE clause like this 
-- SELECT X,Y,Z From SomeTable 
-- WHERE DateField BETWEEN @PreviousMondayToLastSunday AND @LastSunday

答

这将让你的未来和珍贵周五从给定的日期和时间

DECLARE @PREVIOUS int, @dtmStart datetime,@dtmEnd datetime, @NEXT int; 
SET @dtmStart = '12/10/2013'; 
SET @dtmEnd = '12/11/2013'; 

select @PREVIOUS = datepart(dw,@dtmStart) 
WHILE @PREVIOUS <> 6 
BEGIN 
    SET @dtmStart = DATEADD(day , -1 ,@dtmStart) 
    SET @PREVIOUS = datepart(dw,@dtmStart) 
    CONTINUE 
END 
select @dtmStart 

SELECT @NEXT = DATEPART(dw, @dtmEnd) 
    WHILE @NEXT <> 6 
BEGIN 
    SET @dtmEnd = DATEADD(day , 1 ,@dtmEnd) 
    SET @NEXT = datepart(dw,@dtmEnd) 
    CONTINUE 
END 
select @dtmEnd

答

为了得到一个星期日，还是今天，如果今天是星期天，试试这个

DATEADD(day,- (DATEPART(dw,getdate()) + @@DATEFIRST -1) % 7, getdate())

答

这里是获取上一个星期六的日期的代码唉。此方法独立于数据库的设置。

declare @lastSaturday date, 
     @today date, 
     @todayName varchar(20); 

select @todayName = datename(weekday, getdate()), @today = getdate(); 

select 
      case @todayName 
       when 'Saturday' then @today 
       when 'Sunday' then dateadd(day,-1,@today) 
       when 'Monday' then dateadd(day,-2,@today) 
       when 'Tuesday' then dateadd(day,-3,@today) 
       when 'Wednesday' then dateadd(day,-4,@today) 
       when 'Thursday' then dateadd(day,-5,@today) 
       when 'Friday' then dateadd(day,-6,@today) 
      end as LastSaturday;

如何在sql中获取上周的最后一天？

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