Grouping some result sql fields into a common object - sql

Question

I want to understand if it is possible using the sqlalchemy core syntax to get the given data in the same format as with pure SQL.
DB relations
events = Table(
"events",
metadata,
Column("id", Integer(), primary_key=True),
......
Column("location_id", ForeignKey(locations.c.id, ondelete="SET NULL"), nullable=False),
Column("activities_id", ForeignKey(activities.c.id, ondelete="SET NULL"), nullable=False),
)
locations = Table(
"locations",
metadata,
Column("id", Integer(), primary_key=True),
....
....
)
activities = Table(
"activities",
metadata,
Column("id", Integer(), primary_key=True),
....
)
I want to group the fields from the tables locations and activities into a separate group. In plain SQL this query runs like this (using Postgresql)
SELECT json_build_object('id', e.id, 'title', e.title, 'creator', json_agg(c), " \
'activitie', json_agg(a), 'users', jsonb_agg(u), 'location', jsonb_agg(l)) " \
AS event FROM events AS e " \
LEFT JOIN event_users AS eu ON e.id = eu.events_id " \
LEFT JOIN users AS u ON eu.users_id = u.id " \
LEFT JOIN users AS c ON e.creator = c.id " \
LEFT JOIN activities AS a ON e.activities_id = a.id " \
LEFT JOIN locations AS l ON e.location_id = l.id " \
WHERE e.id = {_id} GROUP BY e.id
And the result is
{'event':
'{
"id" : 1, "title" : "Test",
"creator" : [{"id":1,"created_at":"2021-03-03T23:39:23.469751+03:00","email":"test#email","phone":"232323","hashed_password":"sdsdsds","is_active":true}],
"activitie" : [{"id":1,"name":"basketball","is_active":true}],
"users" : [null],
"location" : [{"id": 3, "lat": 54.49142965, "city": "Berlin", "long": 26.9173560217231, "street": "Stephans", "building": "12"}]
}'
}
As you can see each Foreignkey field is grouped under a common key.
Now I'm trying to do something similar on the sqlalchemy core (without the user table)
query = (
select(
[
events.c.id,
events.c.title,
locations.c.city,
locations.c.street,
locations.c.building,
activities.c.name,
]
)
.select_from(
events.join(locations).join(activities)
)
.where(
and_(
events.c.id == pk,
locations.c.id == events.c.location_id,
activities.c.id == events.c.activities_id)
)
.order_by(desc(events.c.created_at))
)
print(query)
ev = dict(await database.fetch_one(query))
And get the result
{'id': 1, 'title': 'Test', 'city':Berlin', street': 'Stephans', 'building': '12', 'name': 'basketball'}
How to group result like ?
{
'id': 1,
'title': 'Test',
'location': [
'city': 'Berlin'
'street': 'Stephans',
'building': '12',
],
'activity': [
'name': 'basketball'
]
}
p.s. sql query with #van's code
SELECT json_build_object(:json_build_object_2, events.id, :json_build_object_3, events.title, :json_build_object_4, json_agg(json_build_object(:json_build_object_5, locations.city, :json_build_object_6, locations.street, :json_build_object_7, locations.building)), :json_build_object_8, json_agg(json_build_object(:json_build_object_9, locations.id, :json_build_object_10, locations.lat, :json_build_object_11, locations.long, :json_build_object_12, locations.city, :json_build_object_13, locations.street, :json_build_object_14, locations.building)), :json_build_object_15, json_agg(json_build_object(:json_build_object_16, activities.name))) AS json_build_object_1

Answer 1

Below query should do the job:
from sqlalchemy import func
# ...
query = (
select(
[
func.json_build_object(
"id",
events.c.id,
"title",
events.c.title,
"location",
func.json_agg(
func.json_build_object(
"city",
locations.c.city,
"street",
locations.c.street,
"building",
locations.c.building,
)
),
"location_all_columns_example",
func.json_agg(func.json_build_object(
*itertools.chain(*[(_.name, _) for _ in locations.c])
)),
"activity",
func.json_agg(
func.json_build_object(
"name",
activities.c.name,
)
),
)
]
)
.select_from(events.join(locations).join(activities))
.where(
and_(
events.c.id == pk,
locations.c.id == events.c.location_id,
activities.c.id == events.c.activities_id,
)
)
.order_by(desc(events.c.created_at))
.group_by(events.c.id) # !!! <- IMPORTANT
)
Please note that you need the group_by clause.