Sql 获取指定日期的前一个星期二（或一周中的任何一天）

我想得到指定日期的前一个星期二（或一周中的任何一天）。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/???

---

使用T-SQL获取上一个星期二（或一周中的任何一天）最合适的方法是什么？

您可以使用

日期部分

获取周数，然后使用以下语句中的

CASE

语句：

SELECT
    testdate,
    CASE DATEPART(dw,testdate)  WHEN 1 THEN DATEADD(dd,-5,testdate) 
                                WHEN 2 THEN DATEADD(dd,-6,testdate) 
                                WHEN 3 THEN DATEADD(dd, 0,testdate) 
                                WHEN 4 THEN DATEADD(dd,-1,testdate) 
                                WHEN 5 THEN DATEADD(dd,-2,testdate) 
                                WHEN 6 THEN DATEADD(dd,-3,testdate) 
                                WHEN 7 THEN DATEADD(dd,-4,testdate)                                     
    END
FROM #temp

SET DATEFIRST 7

根据@jpw comment，您必须在以下内容中设置为7（默认值）：

SELECT
    testdate,
    CASE DATEPART(dw,testdate)  WHEN 1 THEN DATEADD(dd,-5,testdate) 
                                WHEN 2 THEN DATEADD(dd,-6,testdate) 
                                WHEN 3 THEN DATEADD(dd, 0,testdate) 
                                WHEN 4 THEN DATEADD(dd,-1,testdate) 
                                WHEN 5 THEN DATEADD(dd,-2,testdate) 
                                WHEN 6 THEN DATEADD(dd,-3,testdate) 
                                WHEN 7 THEN DATEADD(dd,-4,testdate)                                     
    END
FROM #temp

SET DATEFIRST 7

---

我希望这对你有帮助

SELECT DATEADD(day,- (DATEPART(dw, testdate) + @@DATEFIRST - 3) % 7,testdate) AS Saturday  
from #temp

或

输出格式将低于格式

Saturday
Tuesday  10/06/2015

---

备注：整个查询只是对时间的组合和计算，要操作时间，最有效的方法是使用DATEADD和DATEDIFF：

CREATE PROC FIND_TUESDAY_DATE
(
 @MYDATE DATE
)
AS 
BEGIN
    SELECT CASE 
            WHEN DATENAME(DW,CAST(DATEADD(DAY,0,@MYDATE) AS DATE)) = 'Tuesday' OR DATENAME(DW,CAST(DATEADD(DAY,-7,@MYDATE) AS DATE))  = 'Tuesday' 
            THEN CAST(DATEADD(DAY,-7,@MYDATE) AS DATE)
            WHEN DATENAME(DW,CAST(DATEADD(DAY,-1,@MYDATE) AS DATE) ) = 'Tuesday' THEN CAST(DATEADD(DAY,-1,@MYDATE) AS DATE)
            WHEN DATENAME(DW,CAST(DATEADD(DAY,-2,@MYDATE) AS DATE) ) = 'Tuesday' THEN CAST(DATEADD(DAY,-2,@MYDATE) AS DATE)
            WHEN DATENAME(DW,CAST(DATEADD(DAY,-3,@MYDATE) AS DATE) ) = 'Tuesday' THEN CAST(DATEADD(DAY,-3,@MYDATE) AS DATE)
            WHEN DATENAME(DW,CAST(DATEADD(DAY,-4,@MYDATE) AS DATE) ) = 'Tuesday' THEN CAST(DATEADD(DAY,-4,@MYDATE) AS DATE)
            WHEN DATENAME(DW,CAST(DATEADD(DAY,-5,@MYDATE) AS DATE) ) = 'Tuesday' THEN CAST(DATEADD(DAY,-5,@MYDATE) AS DATE)
            WHEN DATENAME(DW,CAST(DATEADD(DAY,-6,@MYDATE) AS DATE) ) = 'Tuesday' THEN CAST(DATEADD(DAY,-6,@MYDATE) AS DATE)
        END
END
GO

这是您需要的所有代码，它也适用于其他工作日

DECLARE @daystoadd int = 1 -- tuesday


SELECT DATEADD(week, datediff(d, @daystoadd, testdate) / 7, @daystoadd)
FROM #temp

---

如果仍有帮助，请回答一些细节：

declare @given_date datetime = '2015-10-15 00:30'
declare @required_weekday int = 1 -- tuesday

select dateadd(week, datediff(week, 0, @given_date), @required_weekday)

说明：

datediff（周，0，@给定日期）

：返回1900-01-01和给定日期之间的周数

dateadd（周、周数、周数）

：将计算的周数和给定的周数添加到初始日历日期1900-01-01

---

@SalmanA

DATEFIRST

指定一周的第一天。美国英语默认值为7，星期日

DATEPART

返回一个整数，该整数表示指定日期的指定日期部分

DATEADD

返回指定日期，并将指定的数字间隔（带符号整数）添加到该日期的指定日期部分。整个查询只是时间的组合和计算这是我在SOPossible duplicate上看到的最优雅的答案之一