将每周总计转换为每日平均值答案

【问题标题】：Convert Weekly Total to Daily Average将每周总计转换为每日平均值
【发布时间】：2018-01-24 21:30:46
【问题描述】：

我有一张这样的预测表：

Item      ForecastDate     ForecastQty
A123      7/30/17          140
A123      8/6/17           70
A123      8/13/17          70
A123      8/20/17          70
A123      8/27/17          70
A123      9/3/17           45

我需要得出所有日子（包括周末）的每日平均值，所以我的输出看起来像：

Item      ForecastDate     DailyFcstQty
A123      7/30/17          20
A123      7/31/17          20
A123      8/1/17           20
A123      8/2/17           20
A123      8/3/17           20
A123      8/4/17           20
A123      8/5/17           20
A123      8/6/17           20
A123      8/7/17           10
A123      8/8/17           10

等等。

如何将此每周数据（如果重要，则始终为星期日）转换为这些天的每日平均值？

【问题讨论】：

标签： sql sql-server-2008 date

【解决方案1】：

您需要一个日期表，这可以通过 CTE 完成。然后，完全加入以列举您的日期......并进行除法。由于您声明数据中的日期始终是星期日，因此 7 的除数是静态的，dateadd(day,6,t.ForecastDate) 也是如此。

declare @table table (Item varchar(4), ForecastDate date, ForecastQty int)
insert into @table
values
('A123','7/30/17',140),
('A123','8/6/17',70),
('A123','8/13/17',70),
('A123','8/20/17',70),
('A123','8/27/17',70),
('A123','9/3/17',45)

declare @minDate date = (select min(ForecastDate) from @table)
declare @maxDate date = (select max(ForecastDate) from @table)
;with GetDates As  
(  
select @minDate as TheDate              --startdate
from @table
UNION ALL  
select DATEADD(day,1, TheDate) from GetDates  
where TheDate <= @maxDate   --maxdate
)

select distinct * 
into #tempDates 
from GetDates 
option(maxrecursion 0)

select * from #tempDates order by TheDate
select 
    t.Item
    ,d.TheDate
    ,DailyFcstQty = t.ForecastQty / 7
from @table t
full outer join
    #tempDates d on 
    d.TheDate >= t.ForecastDate
    and d.TheDate <= dateadd(day,6,t.ForecastDate)
order by
    t.ForecastDate


drop table #tempDates

+------+------------+--------------+
| Item |  TheDate   | DailyFcstQty |
+------+------------+--------------+
| A123 | 2017-07-30 |           20 |
| A123 | 2017-07-31 |           20 |
| A123 | 2017-08-01 |           20 |
| A123 | 2017-08-02 |           20 |
| A123 | 2017-08-03 |           20 |
| A123 | 2017-08-04 |           20 |
| A123 | 2017-08-05 |           20 |
| A123 | 2017-08-06 |           10 |
| A123 | 2017-08-07 |           10 |
| A123 | 2017-08-08 |           10 |
| A123 | 2017-08-09 |           10 |
| A123 | 2017-08-10 |           10 |
| A123 | 2017-08-11 |           10 |
| A123 | 2017-08-12 |           10 |
| A123 | 2017-08-13 |           10 |
| A123 | 2017-08-14 |           10 |
| A123 | 2017-08-15 |           10 |
| A123 | 2017-08-16 |           10 |
| A123 | 2017-08-17 |           10 |
| A123 | 2017-08-18 |           10 |
| A123 | 2017-08-19 |           10 |
| A123 | 2017-08-20 |           10 |
| A123 | 2017-08-21 |           10 |
| A123 | 2017-08-22 |           10 |
| A123 | 2017-08-23 |           10 |
| A123 | 2017-08-24 |           10 |
| A123 | 2017-08-25 |           10 |
| A123 | 2017-08-26 |           10 |
| A123 | 2017-08-27 |           10 |
| A123 | 2017-08-28 |           10 |
| A123 | 2017-08-29 |           10 |
| A123 | 2017-08-30 |           10 |
| A123 | 2017-08-31 |           10 |
| A123 | 2017-09-01 |           10 |
| A123 | 2017-09-02 |           10 |
| A123 | 2017-09-03 |            6 |
| A123 | 2017-09-04 |            6 |
+------+------------+--------------+

【讨论】：

这很棒。在注释掉语句中间附近的“select * from #tempDates order by TheDate”后，我能够让您的代码返回结果。我仍然无法将它从我的一般示例调整到我的特定数据集，但这取决于我。谢谢！
现在我已经掌握了我的数据，谢谢@scsimon！