Question

我有一张这样的表：

[Date] [hour] [valid] [value] [type]

我需要对23个先前值及其自身（24个值）中的每个小时执行averange，并且仅当我有18个或更多有效值时才这样做。

我用这个查询做了一个小时：

select type, avg(value) 
from table 
where type = 10 
and valid = 1
and ((date = '11/10/2016' and hour <= 22) or
     (date = '11/09/2016' and hour > 22))
group by type
having count(*)>=18;

但我需要在该时间间隔内每小时的日期间隔。有可能吗？

谢谢。

Answer 1

您可以在下面写一下，我假设列valid包含Y和N值。

select type
       ,max(avg1)
from (
select type
      ,avg(value) over (order by trunc(date,'HH') range between interval '24' hour preceding and interval '0' hour following ) as avg1
      ,count(decode(valid,'Y',1)) over (order by trunc(date,'HH') range between interval '24' hour preceding and interval '0' hour following ) as count1
from table
where type = 10 
and valid = 1)
where count1 >= 18
group by type;

Answer 2

抱歉，但与小时分开保存日期并不好。所以我将它合并到with子句中的日期。如果您不希望这样，您可以在我to_date(date || ':' || hour, 'dd/mm/yyyy:hh24')的任何地方使用d并忽略with条款

with dat as (select to_date(date || ':' || hour, 'dd/mm/yyyy:hh24') d, valid, value, type from table)
select d1.d, d1.type, avg(d2.value) 
  from dat d1 cross join dat d2 
  where d1.d < (d2.d + 1) and d1.d >= d2.d and d2.valid = 1
  group by d1.d, d1.type 
  having count(*) >= 18;

我希望它能奏效。如果您提供了任何测试数据，我会测试它。

Oracle按顺序排列的值

2 个答案: