Sql query to json result - sql

Question

I have these tables
destination (id integer, name text)
activity (id integer, name text, destination_id integer)
activity_duration (value numeric, unit text, label text, activity_id integer)
activity_duration_price(amount numeric, currency_id integer, activity_duration_id integer)
Each destination has 1 or more activities, each activity has 1 or more duration and each duration has 1 price.
here is my query
select
d.id destination_id,
d.name destination_name,
count(a.id) as nb_activity,
string_agg('{"id":'||a.id||',"name":"'||a.name||'","pictures":"'||a.pictures||'","meta":{"price":'||amount||',"label":"'||label||'"}}',',') activities
from destination d
left join activity a on (a.destination_id = d.id)
left join activity_duration ad on (activity_id = a.id)
left join activity_duration_price adp on (adp.activity_duration_id = ad.id)
where d.id = any (array[142]) group by d.id, d.name order by nb_activity desc
and this is the result for the string_agg
[{
"id": 2,
"name": "visit tokyo",
"meta": {
"price": 210,
"label": "Adult"
}
}, {
"id": 2,
"name": "visit tokyo",
"pictures": "{}",
"meta": {
"price": 170,
"label": "Children"
}
}]
As you can see, the meta information is there twice for the same item.
I want my result for the string_agg to be like this
[{
"id": 2,
"name": "visit tokyo",
"meta": [{
"price": 210,
"label": "Adult"
},
{
"price": 170,
"label": "Children"
}
]
}]
How can I get this result?

Answer 1

array_agg - to create array of rows
array_to_json - to create JSON array
Then cast the JSON array to text.
SELECT
d.id destination_id,
d.name destination_name,
count(a.id) AS nb_activity,
string_agg('{"id":'||a.id||',"name":"'||a.name||'","pictures":"'||a.pictures||'","meta":'||
(
SELECT array_to_json(array_agg(t))::TEXT --cast to TEXT to concatenate to the string
FROM
(
SELECT
adp.amount,
ad.label
FROM activity a2
LEFT JOIN activity_duration ad ON (ad.activity_id = a2.id)
LEFT JOIN activity_duration_price adp ON (adp.activity_duration_id = ad.id)
) AS t
)
||'}'
,','
) AS activities
FROM destination d
LEFT JOIN activity a ON (a.destination_id = d.id)
WHERE d.id = ANY (ARRAY[142]) GROUP BY d.id, d.name ORDER BY nb_activity DESC