如何在sql中获取上周的最后一天？答案

【问题标题】：How to get last day of last week in sql?如何在sql中获取上周的最后一天？
【发布时间】：2016-10-27 20:09:47
【问题描述】：

如何在sql中获取上周的最后日期？我的意思是使用查询的上周日日期？

【问题讨论】：

【解决方案1】：

不管实际的 DATEFIRST 设置如何，最后一个星期日都可以这样找到：

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

将GETDATE() 替换为参数@date 以获取特定日期之前的最后一个星期日。

【讨论】：

哇，当我上周想出那个模数技巧时，我以为我太聪明了，可以在给定日期之后找到下一个星期五。我向您致敬，先生。对于那里的非数学极客，Andriy 使用余数返回自周日以来的天数。我要竖起大拇指！

【解决方案2】：

上周日（“上周”的结束时间）

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

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

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

解释巫术的示例（使用此示例将 SQL 更改为所需的每周开始和结束日期）

以下示例定位当前一周中的星期几（周日至周六）
如果实际的 END_OF_WEEK 是下周日至周六周，那么您需要 +7 到本周的值。（参见上面的 END_OF_WEEK 示例。）

SQL 下

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...

【讨论】：

【解决方案3】：

这是一篇关于如何做到这一点的精彩文章：

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

您可能想使用@StartOfPrevWeek 变量。

【讨论】：

【解决方案4】：

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

【讨论】：

【解决方案5】：

要获取上一个星期天，或者如果今天是星期天，请试试这个

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

【讨论】：

【解决方案6】：

使用合适的日历表，SQL 更加简单明了。没有巫毒。

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

【讨论】：

【解决方案7】：

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

【讨论】：

【解决方案8】：

这将使您从给定的日期和时间获得下一个宝贵的星期五

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

【讨论】：

【解决方案9】：

这是获取上周六日期的代码。此方法与数据库的设置无关。

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;

【讨论】：

【解决方案10】：

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

【讨论】：

这不会返回正确的日期。现在运行它会返回 4/30，也就是上周六。