Let's say I have two tables:
User_Combination
+--------+----------------+
| id | combination |
+--------+----------------+
| 6 | [1, 2] |
| 9 | [2, 3] |
+--------+----------------+
Colors
+--------+----------------+
| id | color |
+--------+----------------+
| 1 | Blue |
| 2 | Yellow |
| 3 | Green |
+--------+----------------+
I'm trying to join the result of json_array_elements(color)
with the id
of Elements. For example, the result of
select json_array_elements(color) as CombinationID
from User_Combination where id = 6;
is
+-------------------+
| CombinationID |
+-------------------+
| 1 |
| 2 |
+-------------------+
I'm unable to join CombinationID
with Colors.id
. When I try an SQL command such as:
select json_array_elements(article_data) AS articlesInOutfits (color) as CombinationID
from User_Combination uc JOIN Colors co ON co.id = articlesInOutfits;
or
select json_array_elements(article_data) AS articlesInOutfits (color) as CombinationID
from User_Combination uc JOIN Colors co ON co.id = uc.articlesInOutfits;
It says articlesInOutfits
does not exist.
Any advice?
Use unnest()
to get unpacked combinations:
select id, unnest(combination) cid
from user_combination;
id | cid
----+-----
6 | 1
6 | 2
9 | 2
9 | 3
(4 rows)
Use unpacked cids
to join with colors
:
select u.id, color
from (
select id, unnest(combination) cid
from user_combination
) u
join colors c
on cid = c.id;
id | color
----+--------
6 | Blue
6 | Yellow
9 | Yellow
9 | Green
(4 rows)
Use an aggregate function (e.g. json_agg()
) to get joined colors aggregated for a user:
select u.id, json_agg(color)
from (
select id, unnest(combination) cid
from user_combination
) u
join colors c
on cid = c.id
group by 1;
id | json_agg
----+---------------------
9 | ["Yellow", "Green"]
6 | ["Blue", "Yellow"]
(2 rows)
If combination
is of type json
you should use json_array_elements()
in a lateral join:
select u.id, json_agg(color)
from (
select id, cid
from user_combination,
lateral json_array_elements(combination) cid
) u
join colors c
on cid::text::int = c.id
group by 1;