问题描述
我有下表:
PersonID CW_MilesRun PW_MilesRun CM_MilesRun PM_MilesRun
1 15 25 35 45
2 10 20 30 40
3 5 10 15 20
...
我需要将此表拆分为一个垂直表,每个字段都有一个 id(即 CD_MilesRun =1、CW_MilesRun = 2 等),以便我的表看起来像这样:
PersonID TimeID Description C_MilesRun P_MilesRun
1 1 Week 15 25
1 2 Month 35 45
2 1 Week 10 20
2 2 Month 30 40
3 1 Week 5 10
3 2 Month 15 20
在 postgres 中,我会使用类似于:
SELECT
PersonID,unnest(array[1,2]) AS TimeID,unnest(array['Week','Month']) AS "Description",unnest(array["CW_MilesRun","CM_MilesRun"]) C_MilesRun,unnest(array["PW_MilesRun","PM_MilesRun"]) P_MilesRun
FROM myTableHere
;
但是,我无法在雪花中使用类似的功能。有什么想法吗?
解决方法
您可以将 FLATTEN()
与 LATERAL
结合使用以获得您想要的结果,尽管查询完全不同。
with tbl as (select $1 PersonID,$2 CW_MilesRun,$3 PW_MilesRun,$4 CM_MilesRun,$5 PM_MilesRun from values (1,15,25,35,45),(2,10,20,30,40),(3,5,20))
select
PersonID,t.value[0] TimeID,t.value[1] Description,iff(t.index=0,CW_MilesRun,CM_MilesRun) C_MilesRun,iff(t.index=1,PW_MilesRun,PM_MilesRun) P_MilesRun
from tbl,lateral flatten(parse_json('[[1,"Week"],[2,"Month"]]')) t;
PERSONID TIMEID DESCRIPTION C_MILESRUN P_MILESRUN
1 1 "Week" 15 25
1 2 "Month" 35 45
2 1 "Week" 10 20
2 2 "Month" 30 40
3 1 "Week" 5 10
3 2 "Month" 15 20
附言使用 t.*
查看展平后可用的内容(也许很明显。)
您也可以使用 UNPIVOT 和 NATURAL JOIN。
上面的答案很好......就像考虑其他的做事方式一样......你永远不知道它什么时候可能适合你的需求 - 并且让你接触到一些新的很酷的功能。
with cte as (
select
1 PersonID,15 CW_MilesRun,25 PW_MilesRun,35 CM_MilesRun,45 PM_MilesRun
union
select
2 PersonID,10 CW_MilesRun,20 PW_MilesRun,30 CM_MilesRun,40 PM_MilesRun
union
select
3 PersonID,5 CW_MilesRun,10 PW_MilesRun,15 CM_MilesRun,20 PM_MilesRun
)
select * from
(select
PersonID,CW_MilesRun weekly,CM_MilesRun monthly
from
cte
) unpivot (C_MilesRun for description in (weekly,monthly))
natural join
(select * from
(select
PersonID,PW_MilesRun weekly,PM_MilesRun monthly
from
cte
) unpivot (P_MilesRun for description in (weekly,monthly))) f