datetime列をピボットし、時刻別に集計するとNULLが返される

Question

datetime 'myDateTime'列を持つテーブルをクエリし、各 'myDateTime'が何回表示されているかを取得し、 'myDateTime 'を日付として、その日の時間を列として表します。

with CTE1 as (select DATEADD(dd, DATEDIFF(dd, 0, myDateTime), 0) myDate, count(myDateTime) as Counts, 
case when DATEPART(hour,(DATEADD(hh, DATEDIFF(hh, 0, myDateTime), 0))) in (0,1,2,3,4,5,6,7,21,22,23) then '8 PM to 7 AM' 
when DATEPART(hour,(DATEADD(hh, DATEDIFF(hh, 0, myDateTime), 0))) = 8 then '8 AM' 
when DATEPART(hour,(DATEADD(hh, DATEDIFF(hh, 0, myDateTime), 0))) = 9 then '9 AM' 
when DATEPART(hour,(DATEADD(hh, DATEDIFF(hh, 0, myDateTime), 0))) = 10 then '10 AM' 
when DATEPART(hour,(DATEADD(hh, DATEDIFF(hh, 0, myDateTime), 0))) = 11 then '11 AM' 
when DATEPART(hour,(DATEADD(hh, DATEDIFF(hh, 0, myDateTime), 0))) = 12 then '12 PM' 
when DATEPART(hour,(DATEADD(hh, DATEDIFF(hh, 0, myDateTime), 0))) = 13 then '1 PM' 
when DATEPART(hour,(DATEADD(hh, DATEDIFF(hh, 0, myDateTime), 0))) = 14 then '2 PM' 
when DATEPART(hour,(DATEADD(hh, DATEDIFF(hh, 0, myDateTime), 0))) = 15 then '3 PM' 
when DATEPART(hour,(DATEADD(hh, DATEDIFF(hh, 0, myDateTime), 0))) = 16 then '4 PM' 
when DATEPART(hour,(DATEADD(hh, DATEDIFF(hh, 0, myDateTime), 0))) = 17 then '5 PM' 
when DATEPART(hour,(DATEADD(hh, DATEDIFF(hh, 0, myDateTime), 0))) = 18 then '6 PM' 
when DATEPART(hour,(DATEADD(hh, DATEDIFF(hh, 0, myDateTime), 0))) = 19 then '7 PM' 
when DATEPART(hour,(DATEADD(hh, DATEDIFF(hh, 0, myDateTime), 0))) = 20 then '8 PM' 
else 'Error' end as HourOfDay 
from [table1] with(nolock) 
where myDateTime is not null 
and datediff(day, myDateTime, getdate()) <= 10 
group by DATEADD(dd, DATEDIFF(dd, 0, myDateTime), 0), DATEPART(hour,(DATEADD(hh, DATEDIFF(hh, 0, myDateTime), 0))) 
), 
CTE2 as (select myDate, sum(Counts) as Counts, HourOfDay 
from CTE1 
group by myDate, HourOfDay 
--order by HourOfDay desc 
) 
--CTE3 as (select distinct HourOfDay as hod from cte2) 

select myDate, [1], [2], [3], [4], [5], [6], [7], [8], [9], [10], [11], [12], [13], [14] 
from CTE2 
PIVOT 
(
sum(Counts) 
for HourOfDay in ([1], [2], [3], [4], [5], [6], [7], [8], [9], [10], [11], [12], [13], [14]) 
) as piv 

私が残念ながら得ているのは、すべてがnullです。 1-14の数字は午前8時から午後8時までに表示され、その時間枠以外の時間には余分な列が表示されます。私は間違って何をしていますか？

myDate 1 2 3 4 5 6 7 8 9 10 11 12 13 14 
2016-04-13 00:00:00.000 NULL NULL NULL NULL NULL NULL NULL NULL NULL NULL NULL NULL NULL NULL 
2016-04-14 00:00:00.000 NULL NULL NULL NULL NULL NULL NULL NULL NULL NULL NULL NULL NULL NULL 
2016-04-15 00:00:00.000 NULL NULL NULL NULL NULL NULL NULL NULL NULL NULL NULL NULL NULL NULL 

Answer 1

あなたはCTEの内部に設けられた値に従ってクエリであなたのPIVOTの列名を変更し

（列を[1]実際にある[1 PM]、[2] [2 PM]など）

select * 
from CTE2 
PIVOT (sum(Counts) for HourOfDay in ([1 PM], [2 PM], [3 PM], [4 PM], [5 PM], [6 PM] 
      , [7 PM], [8 PM], [9 PM], [10 PM], [11 PM], [12 PM], [13 PM], [14 PM]) 
) as piv