我已经阅读了很多关于如何“转置”或转移数据的问题，但是没有什么能让我看到我想做的事情是否可能。
在Postgres表中给出一组数据，如下所示：
issue valid station value
2013-02-01 2013-02-01 A 1
2013-02-01 2013-02-02 A 1.2
2013-02-01 2013-02-03 A 1.3
2013-02-01 2013-02-04 A 1.0
2013-02-01 2013-02-01 B 2.1
2013-02-01 2013-02-02 B 2.1
2013-02-01 2013-02-03 B 2.4
2013-02-01 2013-02-04 B 2.7
2013-02-01 2013-02-01 C 3.2
2013-02-01 2013-02-02 C 3.7
2013-02-01 2013-02-03 C 3.5
2013-02-01 2013-02-04 C 3.5
我希望能够查询如果我设置issue=2013-02-01会导致：
station val_day1 day1 val_day1 day2 val_day3 day3 val_day4 day4 max_val
A 1 2013-02-01 1.2 2013-02-02 1.3 2013-02-03 1.0 2013-02-04 1.3
B 2.1 2013-02-01 2.1 2013-02-02 2.4 2013-02-03 2.7 2013-02-04 2.7
C 3.2 2013-02-01 3.7 2013-02-02 3.5 2013-02-03 3.5 2013-02-04 3.7
不幸的是，我对自己可以改变的内容非常有限，所以如果可以在一个很棒的SQL查询中解决这个问题。
对于每个问题，数据始终有4天不同的“有效”日期。每天都会增加一个新问题。
这甚至可以尝试在Postgres开始吗？
由于
答案 0 :(得分：1)
您可以使用CASE和MAX组合来转动数据：
select v2.station,
max(day1) as day1,
max(val_day1) as val_day1,
max(day2) as day2,
max(val_day2) as val_day2,
max(day3) as day3,
max(val_day3) as val_day3,
max(day4) as day4,
max(val_day4) as val_day4,
max(greatest(val_day1, val_day2, val_day3, val_day4)) as max_val
from (
select v1.*,
(case when v1.day = 1 then v1.valid else NULL end) as day1,
(case when v1.day = 1 then v1.value else NULL end) as val_day1,
(case when v1.day = 2 then v1.valid else NULL end) as day2,
(case when v1.day = 2 then v1.value else NULL end) as val_day2,
(case when v1.day = 3 then v1.valid else NULL end) as day3,
(case when v1.day = 3 then v1.value else NULL end) as val_day3,
(case when v1.day = 4 then v1.valid else NULL end) as day4,
(case when v1.day = 4 then v1.value else NULL end) as val_day4
from
(select t.valid, t.station, t.value, t.valid - t.issue +1 as day
from so_data t) v1
) v2
group by v2.station
order by v2.station