SQL聚合OVER和PARTITION

Question

所有，

这是我关于Stackoverflow的第一篇文章，所以轻松吧...

我正在使用SQL Server 2008。

我刚开始编写SQL查询，但遇到一个我认为很简单的问题，但是我已经奋斗了2天。 我有一组看起来像这样的数据：

UserId          Duration(Seconds)        Month
1               45                       January
1               90                       January
1               50                       February
1               42                       February
2               80                       January
2               110                      February
3               45                       January
3               62                       January
3               56                       January
3               60                       February

现在，我要编写一个查询，为我提供特定用户的平均值，并将其与该月所有用户的平均值进行比较。 因此，查询用户＃1后的结果数据集将如下所示：

UserId         Duration(seconds)        OrganizationDuration(Seconds)        Month
1              67.5                     63                                   January
1              46                       65.5                                 February

我一直在围绕不同的子查询和按场景分组，似乎没有任何工作。 最近，我一直在尝试OVER和PARTITION BY，但也没有成功。 我最新的查询如下所示：

select Userid, 
       AVG(duration) OVER () as OrgAverage,
       AVG(duration) as UserAverage,
       DATENAME(mm,MONTH(StartDate)) as Month
            from table.name 
            where YEAR(StartDate)=2014
            AND userid=119 
                  GROUP BY MONTH(StartDate), UserId     

该查询在选择列表中用“持续时间”轰炸，因为它没有包含在聚合函数或GROUP BY子句中，因此无效。

请记住，我正在处理大量数据。 我认为我可以使其与CASE语句一起使用，但是我正在寻找一种更简洁，更有效的方式来编写查询（如果可能）。

谢谢！

Answer 1

平均函数中缺少分区子句

OVER ( Partition by MONTH(StartDate)) 

Answer 2

您在这里将两个查询结合在一起：

• 每个用户每月的平均值
• 所有组织平均每月

如果您一次仅要返回一个用户的数据，那么内联选择可能会给您带来欢乐：

SELECT AVG(a.duration) AS UserAvergage,
   (SELECT AVG(b.Duration) FROM tbl b WHERE MONTH(b.StartDate) = MONTH(a.StartDate)) AS OrgAverage 
    ...
    FROM tbl a
    WHERE userid = 119 
    GROUP BY MONTH(StartDate), UserId

注意-在MONTH上使用比较可能会比较慢-使用CTE（通用表表达式）可能会更好

Answer 3

我能够通过自我连接完成它，这可能是一种更好的方法。

Select UserId, AVG(t1.Duration) as Duration, t2.duration as OrgDur, t1.Month 
from #temp t1
inner join (Select Distinct MONTH, AVG(Duration) over (partition by Month) as duration
from #temp) t2 on t2.Month = t1.Month
group by t1.Month, t1.UserId, t2.Duration 
order by t1.UserId, Month desc

这里使用的CTE可能是更好的解决方案，而且绝对更易于阅读

With MonthlyAverage
as 
(
Select MONTH, AVG(Duration) as OrgDur 
from #temp
group by Month
)

Select UserId, AVG(t1.Duration) as Duration, m.duration as OrgDur , t1.Month 
from #temp t1
inner join MonthlyAverage m on m.Month = t1.Month
group by UserId, t1.Month, m.duration

Answer 4

Please try this. It works fine to me.

WITH C1
AS
(
SELECT 
AVG(Duration) AS TotalAvg, 
[Month]
FROM [dbo].[Test]
GROUP BY [Month]
),
C2
AS
(
SELECT Distinct UserID,
AVG(Duration) OVER(PARTITION BY UserID, [Month] ORDER BY UserID) AS DetailedAvg, 
[Month]
FROM [dbo].[Test]
)
SELECT C2.*, C1.TotalAvg
FROM C2 c2 
INNER JOIN C1 c1 ON c1.[Month] = c2.[Month]
ORDER BY c2.UserID, c2.[Month] desc;

Answer 5

您可以在下面用更少的代码尝试。

SELECT Distinct UserID,
AVG(Duration)  OVER(PARTITION BY [Month]) AS TotalAvg,
AVG(Duration) OVER(PARTITION BY UserID, [Month] ORDER BY UserID) AS DetailedAvg, 
[Month]
FROM [dbo].[Test]