jsonb_agg: Avoid objects wrapped with "jsonb_build_object"

Question

I can create JSON objects using jsonb_build_object the way I want them. E.g.

SELECT jsonb_build_object('id', id) FROM (SELECT generate_series(1,3) id) objects;

results in

jsonb_build_object
------------------
{"id": 1}
{"id": 2}
{"id": 3}

But when I want to add them to an array, they are wrapped in an additional object, using the column name as key:

SELECT jsonb_build_object(
  'foo', 'bar',
  'collection', jsonb_agg(collection)
)
FROM (
  SELECT jsonb_build_object('id', id)
  FROM (
    SELECT generate_series(1,3) id
  ) objects
) collection;

results in

{"foo": "bar", "collection": [{"jsonb_build_object": {"id": 1}}, {"jsonb_build_object": {"id": 2}}, {"jsonb_build_object": {"id": 3}}]}

How can I get

{"foo": "bar", "collection": [{"id": 1}, {"id": 2}, {"id": 3}]}

instead?

Answer 1

Use jsonb_agg(collection.jsonb_build_object). You can use aliases too, but the point is that collection refers to the entire row, which has a (single) jsonb_build_object named (by default) column, which is the JSON you want to aggregate.

With simplifying and aliases, you query can be:

SELECT     jsonb_build_object(
             'foo', 'bar',
             'collection', jsonb_agg(js)
           )
FROM       generate_series(1,3) id
CROSS JOIN LATERAL jsonb_build_object('id', id) js;

Notes:

• LATERAL is implicit, I just wrote it for clarity
• aliasing like this in the FROM clause creates a table & a column alias too, with the same name. So it is equivalent to jsonb_build_object('id', id) AS js(js)