作爲一個謎,這是一個解決方案...在實踐中,可根據您的數據的性質可怕執行。注意你的索引,在任何情況下:
create database tmp;
create table t (value float, dt date); -- if you use int, you need to care about rounding
insert into t values (10, '2012-10-30'), (15, '2012-10-29'), (null, '2012-10-28'), (null, '2012-10-27'), (7, '2012-10-26');
select t1.dt, t1.value, t2.dt, t2.value, count(*) cnt
from t t1, t t2, t t3
where
t2.dt >= t1.dt and t2.value is not null
and not exists (
select *
from t
where t.dt < t2.dt and t.dt >= t1.dt and t.value is not null
)
and t3.dt <= t2.dt
and not exists (
select *
from t where t.dt >= t3.dt and t.dt < t2.dt and t.value is not null
)
group by t1.dt;
+------------+-------+------------+-------+-----+
| dt | value | dt | value | cnt |
+------------+-------+------------+-------+-----+
| 2012-10-26 | 7 | 2012-10-26 | 7 | 1 |
| 2012-10-27 | NULL | 2012-10-29 | 15 | 3 |
| 2012-10-28 | NULL | 2012-10-29 | 15 | 3 |
| 2012-10-29 | 15 | 2012-10-29 | 15 | 3 |
| 2012-10-30 | 10 | 2012-10-30 | 10 | 1 |
+------------+-------+------------+-------+-----+
5 rows in set (0.00 sec)
select dt, value/cnt
from (
select t1.dt , t2.value, count(*) cnt
from t t1, t t2, t t3
where
t2.dt >= t1.dt and t2.value is not null
and not exists (
select *
from t
where t.dt < t2.dt and t.dt >= t1.dt and t.value is not null
)
and t3.dt <= t2.dt
and not exists (
select *
from t
where t.dt >= t3.dt and t.dt < t2.dt and t.value is not null
)
group by t1.dt
) x;
+------------+-----------+
| dt | value/cnt |
+------------+-----------+
| 2012-10-26 | 7 |
| 2012-10-27 | 5 |
| 2012-10-28 | 5 |
| 2012-10-29 | 5 |
| 2012-10-30 | 10 |
+------------+-----------+
5 rows in set (0.00 sec)
說明:
- T1是原始表
- T2是與非空值
最少的更大的日期表中的行
- T3成爲之間的所有行,因此我們可以通過其他組和計數
對不起,我不能再清楚不過了。這是混淆對我來說太:-)
+1非常好的問題。它擁有它需要的一切 - 好吧,我從小提琴中推斷出PostgreSQL 9.2。 –