the title may not be very clear so let\'s consider this example (this is not my code, just taking this example to model my request)
I have a table that references itself
Use recursive common table expression. Always starting from the root, use an array of ids to get paths for a given id
in the WHERE
clause.
For id = 1
:
with recursive cte(id, parent, name, ids) as (
select id, parent, name, array[id]
from my_table
where parent is null
union all
select t.id, t.parent, concat(c.name, t.name, '/'), ids || t.id
from cte c
join my_table t on c.id = t.parent
)
select id, name
from cte
where 1 = any(ids) and id <> 1
id | name
----+-----------------------
2 | /home/
5 | /usr/
6 | /usr/local/
3 | /home/user/
4 | /home/user/bin/
(5 rows)
For id = 2
:
with recursive cte(id, parent, name, ids) as (
select id, parent, name, array[id]
from my_table
where parent is null
union all
select t.id, t.parent, concat(c.name, t.name, '/'), ids || t.id
from cte c
join my_table t on c.id = t.parent
)
select id, name
from cte
where 2 = any(ids) and id <> 2
id | name
----+-----------------------
3 | /home/user/
4 | /home/user/bin/
(2 rows)
The question is really interesting. The above query works well but is inefficient as it parses all tree nodes even when we're asking for a leaf. The more powerful solution is a bidirectional recursive query. The inner query walks from a given node to top, while the outer one goes from the node to bottom.
with recursive outer_query(id, parent, name) as (
with recursive inner_query(qid, id, parent, name) as (
select id, id, parent, name
from my_table
where id = 2 -- parameter
union all
select qid, t.id, t.parent, concat(t.name, '/', q.name)
from inner_query q
join my_table t on q.parent = t.id
)
select qid, null::int, right(name, -1)
from inner_query
where parent is null
union all
select t.id, t.parent, concat(q.name, '/', t.name)
from outer_query q
join my_table t on q.id = t.parent
)
select id, name
from outer_query
where id <> 2; -- parameter