4

以下は、SQL を記述する必要があるデータ設定です。

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

これまでのところ、長くてエレガントとはほど遠い以下のクエリがあります。

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;

これをエレガントに解決する方法はありますか？ありがとう。

4

2 に答える 2

3

a を使用cteしてそれを行うことができます。

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;

でテストする SQLfiddle。

...または、通常の階層クエリを使用できます。

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

別の SQLfiddle。

于 2013-06-28T11:46:02.510 に答える
3

connect_by_root を使用できます

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
;
于 2013-06-28T11:39:52.123 に答える