Postgresql 如何创建递归cte查询，将父ID和祖父母ID推送到一个数组中
我正在尝试创建一个postgresql表。这是我的cte，我在这里插入值Postgresql 如何创建递归cte查询，将父ID和祖父母ID推送到一个数组中,postgresql,common-table-expression,recursive-query,Postgresql,Common Table Expression,Recursive Query,我正在尝试创建一个postgresql表。这是我的cte，我在这里插入值 BEGIN; CREATE TABLE section ( id SERIAL PRIMARY KEY, parent_id INTEGER REFERENCES section(id) DEFERRABLE, name TEXT NOT NULL UNIQUE ); SET CONSTRAINTS ALL DEFERRED; INSERT INTO section VALUES (1, NULL, 'anim
BEGIN;
CREATE TABLE section (
id SERIAL PRIMARY KEY,
parent_id INTEGER REFERENCES section(id) DEFERRABLE,
name TEXT NOT NULL UNIQUE );
SET CONSTRAINTS ALL DEFERRED;
INSERT INTO section VALUES (1, NULL, 'animal');
INSERT INTO section VALUES (2, NULL, 'mineral');
INSERT INTO section VALUES (3, NULL, 'vegetable');
INSERT INTO section VALUES (4, 1, 'dog');
INSERT INTO section VALUES (5, 1, 'cat');
INSERT INTO section VALUES (6, 4, 'doberman');
INSERT INTO section VALUES (7, 4, 'dachshund');
INSERT INTO section VALUES (8, 3, 'carrot');
INSERT INTO section VALUES (9, 3, 'lettuce');
INSERT INTO section VALUES (10, 11, 'paradox1');
INSERT INTO section VALUES (11, 10, 'paradox2');
SELECT setval('section_id_seq', (select max(id) from section));
WITH RECURSIVE last_run(parent_id, id_list, name_list) AS (
???
SELECT id_list, name_list
FROM last_run ???
WHERE ORDER BY id_list;
ROLLBACK;
id_list | name_list
---------+------------------------
{1} | animal
{2} | mineral
{3} | vegetable
{4,1} | dog, animal
{5,1} | cat, animal
{6,4,1} | doberman, dog, animal
{7,4,1} | dachshund, dog, animal
{8,3} | carrot, vegetable
{9,3} | lettuce, vegetable
{10,11} | paradox1, paradox2
{11,10} | paradox2, paradox1
您可以在单个查询中使用多个递归CTE：一个用于有效树，另一个用于悖论：
with recursive
cte as (
select *, array[id] as ids, array[name] as names
from section
where parent_id is null
union all
select s.*, s.id||c.ids, s.name||c.names
from section as s join cte as c on (s.parent_id = c.id)),
paradoxes as (
select *, array[id] as ids, array[name] as names
from section
where id not in (select id from cte)
union all
select s.*, s.id||p.ids, s.name||p.names
from section as s join paradoxes as p on (s.parent_id = p.id)
where s.id <> all(p.ids) -- To break loops
)
select * from cte
union all
select * from paradoxes;