我从下表中看到了很多关于转置的问题......
scanid | region | volume
-------------------------
1 A 34.4
1 B 32.1
1 C 29.1
2 A 32.4
2 B 33.2
2 C 35.6
到这张桌子。
scanid | A_volume | B_volume | C_volume
----------------------------------------
1 34.4 32.1 29.1
2 32.4 33.2 35.6
然而,我需要做反过来,并且无法试图解决这个问题。有人可以帮忙吗?
谢谢。
答案 0 :(得分:2)
目前尚不清楚如何恢复" A"," B"," C"值,所以我只是添加它们
制备
t=# create table s188 (scanid int,a float, b float,c float);
CREATE TABLE
t=# insert into s188 select 1,2,3,4;
INSERT 0 1
t=# insert into s188 select 2,12,13,14;
INSERT 0 1
t=# select * from s188;
scanid | a | b | c
--------+----+----+----
1 | 2 | 3 | 4
2 | 12 | 13 | 14
(2 rows)
选择
t=# with a as (
select scanid,unnest(array[a,b,c]) from s188
)
select scanid,chr((row_number() over (partition by scanid))::int + 64),unnest
from a;
scanid | chr | unnest
--------+-----+--------
1 | A | 2
1 | B | 3
1 | C | 4
2 | A | 12
2 | B | 13
2 | C | 14
(6 rows)
来自a_horse_with_no_name
的更整洁的解决方案t=# with a as (
select scanid, x.*
from s188, unnest(array[a,b,c]) with ordinality as x(volume,idx)
)
select scanid,
chr(idx::int + 64) as region,
volume
from a;
scanid | region | volume
--------+--------+--------
1 | A | 2
1 | B | 3
1 | C | 4
2 | A | 12
2 | B | 13
2 | C | 14
(6 rows)
答案 1 :(得分:1)
您可以使用UNION子句非常简单地执行此操作:
Select Scan_ID, 'A' as Region, A_Volume as volume
union all
Select Scan_ID, 'B' as Region, B_Volume as volume
union all
Select Scan_ID, 'C' as Region, C_Volume as volume