Gerichte acyclische grafiek:vind alle paden van een specifiek knooppunt

WITH RECURSIVE parent_line(path, id) AS (
    SELECT ARRAY[(row_number() OVER (PARTITION BY CP.idc))::integer], C.id
    FROM categorias C JOIN categ_parents CP ON C.id = CP.idparent
    WHERE CP.idc = 36
    UNION ALL
    SELECT PL.path || (row_number() OVER (PARTITION BY PL.id))::integer,
        C.id
    FROM categorias C
    JOIN categ_parents CP ON C.id = CP.idparent
    JOIN parent_line PL on CP.idc = PL.id
),
real_parent_line(path, chainid, id) AS (
    SELECT PL.path, (row_number() OVER (PARTITION BY PL.id)),
        PL.id
    FROM parent_line PL
    WHERE PL.id IN (
        SELECT id
        FROM categorias C LEFT JOIN categ_parents CP
            ON (C.id = CP.idc)
        WHERE CP.idc IS NULL
    )
    UNION ALL
    SELECT PL.path, chainid, PL.id
    FROM parent_line PL, real_parent_line RPL
    WHERE array_upper(PL.path,1) + 1 = array_upper(RPL.path,1)
        AND PL.path = RPL.path[1:(array_upper(RPL.path,1)-1)]
)
SELECT array_accum(id) AS ids
FROM real_parent_line RPL
GROUP BY chainid;

De eerste WITH clausule geeft dit:

path             | id
------------------------
"{1}"              31
"{1,1}"            22
"{1,2}"            30
"{1,1,1}"          20
"{1,1,2}"          8
"{1,2,1}"          20
"{1,1,2,1}"        5
"{1,1,1,1}"        1
"{1,2,1,2}"        1
"{1,1,2,1,1}"      2
"{1,1,2,1,1,1}"    1
"{1,1,2,1,1,2}"    20
"{1,1,2,1,1,2,1}"  1

Met dank aan #[email protected] voor wat hulp.