SQL의 복잡한 계층 적 쿼리

Question

다음은 SQL을 작성해야하는 데이터 설정입니다.

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 

나는 길고 우아한에서 멀리 지금까지 아래 쿼리가 있습니다.

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; 

이 방법을 우아하게 해결할 수 있습니까? 감사.

Answer 1

당신은 그것을 할 수있는 cte을 사용할 수 있습니다 connect_by_root 사용할 수 있습니다;

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.

... 또는 일반 계층 쿼리를 사용할 수 있습니다.

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.

Answer 2

당신이

select root, sum(val) 
from 
(select chil, connect_by_root par root 
    from parchil 
    connect by par = prior chil 
    start with par in (select par from parchil) 
    union all 
    select par, par from parchil 
) 
, subval 
where 
    sub=chil 
group by root 
order by root 
;