以下是需要编写 SQL 的数据设置。
Table:parchil
par chil
--------------------
E1 E2
E2 E3
E3 E4
E5 E6
E7 E8
Table:subval
sub val
--------------------
E1 10
E2 70
E3 30
E4 40
E5 60
E6 20
E7 50
Expected result:
sub val
--------------------
E1 150
E2 140
E3 70
E4 40
E5 80
E6 20
E7 50
到目前为止，我有以下查询，它冗长且远非优雅。
select a.par,sum(b.val) from
(select 'E1' as par,'E1' as chil from dual
union all
select
'E1' as par, chil
from
parchil
start with par='E1'
connect by prior chil=par
union all
select 'E2' as par,'E2' as chil from dual
union all
select
'E2' as par, chil
from
parchil
start with par='E2'
connect by prior chil=par
union all
select 'E3' as par,'E3' as chil from dual
union all
select
'E3' as par, chil
from
parchil
start with par='E3'
connect by prior chil=par
union all
select 'E4' as par,'E4' as chil from dual
union all
select
'E4' as par, chil
from
parchil
start with par='E4'
connect by prior chil=par
union all
select 'E5' as par,'E5' as chil from dual
union all
select
'E5' as par, chil
from
parchil
start with par='E5'
connect by prior chil=par
union all
select 'E6' as par,'E6' as chil from dual
union all
select
'E6' as par, chil
from
parchil
start with par='E6'
connect by prior chil=par
union all
select 'E7' as par,'E7' as chil from dual
union all
select
'E7' as par, chil
from
parchil
start with par='E7'
connect by prior chil=par
) a,
subval b
where
a.chil=b.sub
group by a.par
order by a.par;
有没有办法优雅地解决这个问题？谢谢。
最佳答案
您可以使用 cte 来做到这一点；
WITH cte(sub,val,par,chil, lev) AS (
SELECT s.sub, s.val, p.par, p.chil, 1
FROM subval s LEFT JOIN parchil p ON s.sub=p.par
UNION ALL
SELECT s.sub, s.val+c.val, p.par, p.chil, lev + 1
FROM subval s LEFT JOIN parchil p ON s.sub=p.par
JOIN cte c ON c.sub=p.chil
)
SELECT c1.sub,c1.val FROM cte c1
LEFT JOIN cte c2
ON c1.sub=c2.sub
AND c1.lev < c2.lev
WHERE c2.sub IS NULL
ORDER BY sub;
An SQLfiddle to test with 。
...或者您可以使用常规的分层查询；
SELECT root, SUM(val) val
FROM
(
SELECT CONNECT_BY_ROOT sub root, val
FROM subval s
LEFT JOIN parchil p ON s.sub = p.par
CONNECT BY sub = PRIOR chil
)
GROUP BY root
ORDER BY root
Another SQLfiddle 。
关于sql - SQL 中的复杂分层查询，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/17363531/