Question

Avoir la my_tabe tableau suivant:

M01 |   1
M01 |   2
M02 |   1

Je veux requête dessus afin d'obtenir:

M01 |   1,2
M02 |   1

J'ai réussi à se rapprocher en utilisant la requête suivante:

with my_tabe as
(
    select 'M01' as scycle, '1' as sdate from dual union
    select 'M01' as scycle, '2' as sdate from dual union
    select 'M02' as scycle, '1' as sdate from dual
)
SELECT scycle, ltrim(sys_connect_by_path(sdate, ','), ',')
FROM
(
    select scycle, sdate, rownum rn
    from my_tabe
    order by 1 desc
)
START WITH rn = 1
CONNECT BY PRIOR rn = rn - 1

Cédant:

SCYCLE      |   RES
M02         |   1,2,1
M01         |   1,2

Ce qui est faux. Il est semble que je suis proche, mais je crains que je ne pas ce à l'étape suivante ...

Des conseils?

Était-ce utile?

La solution

Vous devez restreindre votre connect by à la même valeur scycle, et compte également le nombre de correspondances et filtre que pour éviter de voir les résultats intermédiaires.

with my_tabe as
(
    select 'M01' as scycle, '1' as sdate from dual union
    select 'M01' as scycle, '2' as sdate from dual union
    select 'M02' as scycle, '1' as sdate from dual
)
select scycle, ltrim(sys_connect_by_path(sdate, ','), ',')
from
(
    select distinct sdate,
        scycle,
        count(1) over (partition by scycle) as cnt,
        row_number() over (partition by scycle order by sdate) as rn
    from my_tabe
)
where rn = cnt
start with rn = 1
connect by prior rn + 1 = rn
and prior scycle = scycle
/

SCYCLE LTRIM(SYS_CONNECT_BY_PATH(SDATE,','),',')
------ -----------------------------------------
M01    1,2
M02    1

Si vous êtes 11g, vous pouvez utiliser le haut- LISTAGG fonction à la place:

with my_tabe as
(
    select 'M01' as scycle, '1' as sdate from dual union
    select 'M01' as scycle, '2' as sdate from dual union
    select 'M02' as scycle, '1' as sdate from dual
)
select scycle, listagg (sdate, ',') 
within group (order by sdate) res
from my_tabe
group by scycle
/ 

Les deux approches (et d'autres) sont affichés .

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