Sal*_*n A 7 sql t-sql sql-server datetime sql-server-2008
我希望在指定日期之前的星期二(或任何特定日期)获得.以下是周二的示例输入和预期输出:
CREATE TABLE #temp(testdate DATETIME);
INSERT INTO #temp(testdate) VALUES
('2015-10-06 01:15'), -- Tue -> Tue 2015-10-06 00:00
('2015-10-07 04:30'), -- Wed -> Tue 2015-10-06 00:00
('2015-10-08 00:30'), -- Thu -> Tue 2015-10-06 00:00
('2015-10-09 21:00'), -- Fri -> Tue 2015-10-06 00:00
('2015-10-10 19:00'), -- Sat -> Tue 2015-10-06 00:00
('2015-10-11 01:15'), -- Sun -> Tue 2015-10-06 00:00
('2015-10-12 13:00'), -- Mon -> Tue 2015-10-06 00:00
('2015-10-13 18:45'), -- Tue -> Tue 2015-10-13 00:00
('2015-10-14 12:15'), -- Wed -> Tue 2015-10-13 00:00
('2015-10-15 10:45'), -- Thu -> Tue 2015-10-13 00:00
('2015-10-16 04:30'), -- Fri -> Tue 2015-10-13 00:00
('2015-10-17 12:15'), -- Sat -> Tue 2015-10-13 00:00
('2015-10-18 00:30'), -- Sun -> Tue 2015-10-13 00:00
('2015-10-19 10:45'), -- Mon -> Tue 2015-10-13 00:00
('2015-10-20 01:15'), -- Tue -> Tue 2015-10-20 00:00
('2015-10-21 23:45'), -- Wed -> Tue 2015-10-20 00:00
('2015-10-22 21:00'), -- Thu -> Tue 2015-10-20 00:00
('2015-10-23 18:45'), -- Fri -> Tue 2015-10-20 00:00
('2015-10-24 06:45'), -- Sat -> Tue 2015-10-20 00:00
('2015-10-25 06:45'), -- Sun -> Tue 2015-10-20 00:00
('2015-10-26 04:30'); -- Mon -> Tue 2015-10-20 00:00
DECLARE @clampday AS INT = 3; -- Tuesday
SELECT -- DATEADD/DATEPART/@clampday/???
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使用T-SQL获取上周二(或任何一周)的最合适方法是什么?
我希望这能帮到您
SELECT DATEADD(day,- (DATEPART(dw, testdate) + @@DATEFIRST - 3) % 7,testdate) AS Saturday
from #temp
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要么
SELECT DATENAME(weekday,DATEADD(day,- (DATEPART(dw, testdate) + @@DATEFIRST - 3) % 7,testdate)) +' '+
CONVERT(nvarchar,DATEADD(day,- (DATEPART(dw, testdate) + @@DATEFIRST - 3) % 7,testdate),101) AS Saturday
from @temp
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输出将在下面的甲酸盐中
Saturday
Tuesday 10/06/2015
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备注:整个查询只是组合和计算DATEFIRST,DATEPART和DATEADD来操纵时间