我遇到周数问题。客户周从星期二开始,因此在星期一结束。所以我做了:
Set DateFirst 2
然后我使用
DateAdd(ww,@WeeksToShow, Date)
偶尔会给我8周的信息。我认为这是因为它过去了一年,但我不知道如何解决它。
如果我这样做:
(DatePart(dy,Date) / 7) - @WeeksToShow
然后它的效果会更好,但显然不会影响前几年,因为它只是减去数字。
编辑:
我目前的SQL(如果它没有任何数据就有帮助)
Set DateFirst 2
Select
DATEPART(yyyy,SessionDate) as YearNo,
DATEPART(ww,SessionDate) as WeekNo,
DATEADD(DAY, 1 - DATEPART(WEEKDAY, SessionDate + SessionTime), CAST(SessionDate +SessionTime AS DATE)) [WeekStart],
DATEADD(DAY, 7 - DATEPART(WEEKDAY, SessionDate + SessionTime), CAST(SessionDate + SessionTime AS DATE)) [WeekEnd],
DateName(dw,DATEADD(DAY, 7 - DATEPART(WEEKDAY, SessionDate + SessionTime), CAST(SessionDate + SessionTime AS DATE))) as WeekEndName,
Case when @ConsolidateSites = 1 then 0 else SiteNo end as SiteNo,
Case when @ConsolidateSites = 1 then 'All' else CfgSites.Name end as SiteName,
GroupNo,
GroupName,
DeptNo,
DeptName,
SDeptNo,
SDeptName,
PluNo,
PluDescription,
SUM(Qty) as SalesQty,
SUM(Value) as SalesValue
From
PluSalesExtended
Left Join
CfgSites on PluSalesExtended.SiteNo = CfgSites.No
Where
Exists (Select Descendant from DescendantSites where Parent in (@SiteNo) and Descendant = PluSalesExtended.SiteNo)
AND (DATEPART(WW,SessionDate + SessionTime) !=DATEPART(WW,GETDATE()))
AND SessionDate + SessionTime between DATEADD(ww,@NumberOfWeeks * -1,@StartingDate) and @StartingDate
AND TermNo = 0
AND PluEntryType <> 4
Group by
DATEPART(yyyy,SessionDate),
DATEPART(ww,SessionDate),
DATEADD(DAY, 1 - DATEPART(WEEKDAY, SessionDate + SessionTime), CAST(SessionDate +SessionTime AS DATE)),
DATEADD(DAY, 7 - DATEPART(WEEKDAY, SessionDate + SessionTime), CAST(SessionDate + SessionTime AS DATE)),
Case when @ConsolidateSites = 1 then 0 else SiteNo end,
Case when @ConsolidateSites = 1 then 'All' else CfgSites.Name end,
GroupNo,
GroupName,
DeptNo,
DeptName,
SDeptNo,
SDeptName,
PluNo,
PluDescription
order by WeekEnd
答案 0 :(得分:2)
这里有两个问题,第一个是我怀疑你将8周的数据定义为DATEPART(WEEK
有8个不同的值,在这种情况下你可以通过查看什么来复制问题的根本原因ISO将定义为2015年的第一周:
SET DATEFIRST 2;
SELECT Date, Week = DATEPART(WEEK, Date)
FROM (VALUES
('20141229'), ('20141230'), ('20141231'), ('20150101'),
('20150102'), ('20150103'), ('20150104')
) d (Date);
给出了:
Date Week
-----------------
2014-12-29 52
2014-12-30 53
2014-12-31 53
2015-01-01 1
2015-01-02 1
2015-01-03 1
2015-01-04 1
因此,虽然您只有7天,但您有3个不同的周数。问题是DATEPART(WEEK
是一个非常简单的函数,并且只返回自一年的第一天起经过的周边界数,更好的函数将是ISO_WEEK
,因为这会考虑年份边界很好地:
SET DATEFIRST 2;
SELECT Date, Week = DATEPART(ISO_WEEK, Date)
FROM (VALUES
('20141229'), ('20141230'), ('20141231'), ('20150101'),
('20150102'), ('20150103'), ('20150104')
) d (Date);
给出了:
Date Week
-----------------
2014-12-29 1
2014-12-30 1
2014-12-31 1
2015-01-01 1
2015-01-02 1
2015-01-03 1
2015-01-04 1
问题是,这并没有考虑星期二开始的星期,因为ISO周从星期一到星期日运行,你可以稍微调整你的使用量来获得前一天的星期数:
SET DATEFIRST 2;
SELECT Date, Week = DATEPART(ISO_WEEK, DATEADD(DAY, -1, Date))
FROM (VALUES
('20141229'), ('20141230'), ('20141231'), ('20150101'),
('20150102'), ('20150103'), ('20150104')
) d (Date);
哪会给:
Date Week
-----------------
2014-12-29 52
2014-12-30 1
2014-12-31 1
2015-01-01 1
2015-01-02 1
2015-01-03 1
2015-01-04 1
因此,12月29日星期一被确认为上一周。问题是内置函数没有ISO_YEAR
,因此您需要定义自己的函数。这是一个相当简单的函数,即使如此我几乎从不创建标量函数,因为它们执行得非常糟糕,而是使用内联表值函数,所以为此我会使用:
CREATE FUNCTION dbo.ISOYear (@Date DATETIME)
RETURNS TABLE
AS
RETURN
( SELECT IsoYear = DATEPART(YEAR, @Date) +
CASE
-- Special cases: Jan 1-3 may belong to the previous year
WHEN (DATEPART(MONTH, @Date) = 1 AND DATEPART(ISO_WEEK, @Date) > 50) THEN -1
-- Special case: Dec 29-31 may belong to the next year
WHEN (DATEPART(MONTH, @Date) = 12 AND DATEPART(ISO_WEEK, @Date) < 45) THEN 1
ELSE 0
END
);
只需要使用子查询,但额外的输入在性能方面是值得的:
SET DATEFIRST 2;
SELECT Date,
Week = DATEPART(ISO_WEEK, DATEADD(DAY, -1, Date)),
Year = (SELECT ISOYear FROM dbo.ISOYear(DATEADD(DAY, -1, Date)))
FROM (VALUES
('20141229'), ('20141230'), ('20141231'), ('20150101'),
('20150102'), ('20150103'), ('20150104')
) d (Date);
或者您可以使用CROSS APPLY
:
SET DATEFIRST 2;
SELECT Date,
Week = DATEPART(ISO_WEEK, DATEADD(DAY, -1, Date)),
Year = y.ISOYear
FROM (VALUES
('20141229'), ('20141230'), ('20141231'), ('20150101'),
('20150102'), ('20150103'), ('20150104')
) d (Date)
CROSS APPLY dbo.ISOYear(d.Date) y;
给出了:
Date Week Year
---------------------------
2014-12-29 52 2014
2014-12-30 1 2015
2014-12-31 1 2015
2015-01-01 1 2015
2015-01-02 1 2015
2015-01-03 1 2015
2015-01-04 1 2015
即使使用这种方法,只需在6周前获得一个日期,如果您使用的日期不是星期二,那么您最终将会持续7周,因为您将有5个星期的整周,并且在开始的一个星期这是第二个问题。所以你需要确保你的开始日期是星期二。以下内容将于7周前发布:
SELECT CAST(DATEADD(DAY, 1 - DATEPART(WEEKDAY, GETDATE()), DATEADD(WEEK, -6, GETDATE())) AS DATE);
在this answer中更好地解释了这个逻辑,以下是将在本周开始的部分(基于你的datefirst设置):
SELECT DATEADD(DAY, 1 - DATEPART(WEEKDAY, GETDATE()), GETDATE());
然后我所做的就是用GETDATE()
替换第二个DATEADD(WEEK, -6, GETDATE())
以便它在6周前获得一周的开始,然后只有一个演员去除时间元素从它。
答案 1 :(得分:0)
这将使您从星期二开始的当前周+前几周开始:
WHERE dateadd(week, datediff(d, 0, getdate()-1)/7 - 4, 1) <= yourdatecolumn
这将显示示例:
DECLARE @wks int = 6 -- Weeks To Show
SELECT
dateadd(week, datediff(d, 0, getdate()-1)/7 - 4, 1) tuesday5weeksago,
dateadd(week, datediff(d, 0, getdate()-1)/7 - 5, 1) tuesday6weeksago,
dateadd(week, datediff(d, 0, getdate()-1)/7 - 6, 1) tuesday7weeksago,
dateadd(week, datediff(d, 0, getdate()-1)/7 - @wks + 1, 1) tuesdaydynamicweeksago
结果:
tuesday5weeksago tuesday6weeksago tuesday7weeksago tuesdaydynamicweeksago
2015-01-27 2015-01-20 2015-01-13 2015-01-20