用枢轴计算平均值

Question

我试图在枢轴的帮助下找到平均值，但无法找到正确的解决方案。

以下是我的疑问：

 select branch, ISNULL([11:00], 0) as [11:00],ISNULL([11:15], 0) as 
 [11:15],ISNULL([11:30], 0) as [11:30], ISNULL([11:45], 0) as [11:45], 
 ISNULL([12:00], 0) as [12:00]

from 
( 
select  b.branchname
       ,convert(varchar(5), intervals.interval_start_time, 108) 
       ,sum(b.ordercount) ordercounts
from branch b cross apply dbo.getDate15MinInterval(CAST(b.TransactionDate 
 as date)) as intervals
 where b.TransactionDate >= interval_start_time and b.TransactionDate <= 
 interval_end_time
 and CAST(TransactionDate AS date) IN ('2017-07-01','2017-07-08')

group by DATEPART(WEEKDAY,TransactionDate),b.branchname,intervals.interval_start_time,intervals.interval_end_time
 ) t 
pivot ( avg(ordercounts) for interval_start_time in ( [11:00], [11:15] , 
[11:30], [11:45], [12:00])) as p

我原来的表是：

以上查询的结果是：

预期结果：

对于15分钟间隔查询，请参考我原来的帖子：
Group data by interval of 15 minutes and use cross tab

Answer 1

SQL Server对整数执行整数运算操作。问题是这是一个整数：

  sum(b.ordercount) as ordercounts

（推测）。

因此，只需将其转换为浮动/固定点数即可。我通常只乘以1.0：

  sum(b.ordercount)*1.0 as ordercounts

但如果您愿意，可以更具体地说明您的类型。

Answer 2

尝试施放以浮动 -

AVG(CAST(ordercounts AS FLOAT))

SUM(CAST(b.ordercount AS FLOAT)) AS ordercounts

Answer 3

可以这样做：

select branchname, [dayname], ISNULL([11:00], 0) as [11:00], AVG(CAST([11:00] as float)) over() [Avg_11:00]
from
(
    select branchname, [dayname], ISNULL([11:00], 0) as [11:00], ISNULL([11:15], 0) as [13:15], ISNULL([11:30], 0) as [11:30], ISNULL([11:45], 0) as [11:45]
    from 
    ( 
        select intervals.[dayname]
             , b.branchname
             , convert(varchar(5), intervals.interval_start_time, 108) interval_start_time -- for hh:mm format
             , sum(b.ordercount) ordercount
        from branch b cross apply dbo.getDate15MinIntervals(CAST(b.TransactionDate as date)) as intervals
        where b.transactiondate between interval_start_time and interval_end_time
        group by intervals.[dayname], b.branchname, intervals.interval_start_time, intervals.interval_end_time
    ) t 
    pivot ( sum(ordercount) for interval_start_time in ( [11:00], [11:15] , [11:30], [11:45] )) as p
) t
group by branchname, [dayname], [11:00]

AVG OVER()自SQL Server 2008起有效。 在这个例子中，我只使用了一个间隔，但你可以将它扩展到你需要的所有间隔。

我尝试了昨天的答案中的一些样本数据，它返回的值如下：

快乐的编码！ ：）