I have some resources in my database that are inherited to their sub-resources. I need to be able to get also the inherited resources when I query a resource. I have a field called path
which I'm planning to use. The path
always contains the full path of all resources related to the resource we are currently handling.
Example:
+-----------------------------------------+
| id | res_id | path |
|-----------------------------------------|
| 1 | res_1 | res_1 |
| 2 | res_1.1 | res_1.res_1.1 |
| 3 | res_1.2 | res_1.res_1.2 |
| 4 | res_1.1.1 | res_1.res_1.1.res_1.1.1|
+-----------------------------------------+
If I query the res_1.1
, I'd also have to get the res_1
because it is the parent of res_1.1
. And if I get the res_1.1.1
, I'd have to get also rows 1 and 2 because they are included in the path of the res_1.1.1
. Would love some advice how to do this with Postgres. I'm also using sqlmodel
to write queries if that's important information.
EDIT. My apologies for the vague introduction, the parameter path
is already a sqlalchemy Ltree -field. I hope this makes things a bit simpler?
ltree
Turns out you are working with the additional module ltree
. And the dot (.
) is your unambiguous delimiter. Assuming path
is type ltree
(you didn't clarify, yet) simplifies the task. There are dedicated operators like:
ltree @> ltree
→boolean
Is left argument an ancestor of right (or equal)?
So:
SELECT t.*
FROM tbl_ltree t1
JOIN tbl_ltree t ON t.path @> t1.path
WHERE t1.res_id = 'res_1_1_1'; -- your search term here
ltree
also provides operator classes for a GiST index:
CREATE INDEX tbl_ltree_path_gist_idx ON tbl USING GIST (path);
.. which can be used by the above query. Plus, you'd have another B-tree index on res_id
.
res_id
is completely redundant and can be ignored or even dropped. We can work with another ltree operator on path
instead:
ltree ~ lquery
→boolean
Does ltree match lquery?
SELECT *
FROM tbl_ltree
WHERE path @> (SELECT path FROM tbl_ltree WHERE path ~ '*.res_1_1_1'::lquery);
db<>fiddle here
Only needs the GiST index.
text
While working with text
instead of ltree
(you didn't clarify, yet), here is one of many ways:
SELECT t.*
FROM (
SELECT unnest(string_to_array(path, '.')) AS res_id
FROM tbl_txt
WHERE res_id = 'res_1_1_1' -- your search term here
) t1
JOIN tbl_txt t USING (res_id);
db<>fiddle here
The subquery t1
splits path
into its building blocks. Then join to another instance of tbl_txt
.