Count other documents in the same table matching the condition - couchbase

Question

I've got this query:
SELECT
id,
(
SELECT COUNT(*)
FROM default t USE KEYS default.id
WHERE t.p_id=default.id
) as children_count
FROM default
WHERE 1
I expect this:
[
{
"children_count": 5,
"id": "1"
},
...
]
But i got this:
[
{
"children_count": [
{
"$1": 0
}
],
"id": "1"
},
...
]
What am I doing wrong? I've googled this, but I can't find any clear explaination of count subqueries in N1QL, so any links to documentation will be highly appreciated.
UPD:
I've updated my code according to #prettyvoid answer. I've also created minimal example bucket to demonstrate the problem.
SELECT
id,
(
SELECT COUNT(*) as count
FROM test t USE KEYS "p.id"
WHERE t.p_id=p.id
)[0].count as children_count
FROM test p
WHERE 1
The result is following:
[
{
"children_count": 0,
"id": 1
},
{
"children_count": 0,
"id": 2
},
{
"children_count": 0,
"id": 3
}
]

Answer 1

Any select statement will yield you an array with objects inside, that's normal. If you want to get your expected result then scope-in to the count object inside the array [0]
edit: Following query will do what you want, I'm unsure if there is a better way though
SELECT
id,
(
SELECT COUNT(*) as count
FROM default t USE KEYS (SELECT RAW meta().id from default)
WHERE t.p_id=p.id
)[0].count as children_count
FROM default p

Answer 2

It's important to note that in Couchbase, the very fastest way to retrieve a document is via its document key. Using a GSI index is workable, but slower. And as with most other databases, it is best to avoid a full scan. You say you can make id the same as the document key, so I will assume that is the case, so that I can use p_id in the on keys clause.
Is it o.k. to only list documents with a non-zero number of children? In that case, you can write this as an aggregation query where you join each child to its parent, and group by the parent id (note my bucket is called default):
select p.id, count(*) as children_count
from default c join default p on keys c.p_id
group by p.id;
If you need to include documents with zero children, you need to UNION with a query that finds those documents as well. In this case we know that:
select raw array_agg(distinct(p_id)) from default where p_id is not null
will give us an array of parent IDs, so we can get the ids not in the list with:
select id, 0 as children_count
from default p
where not array_contains(
(select raw array_agg(distinct(p_id)) from default where p_id is not null)[0],id);
So, if we UNION the two:
(select p.id, count(*) as children_count
from default c join default p on keys c.p_id
group by p.id)
UNION
(select id, 0 as children_count
from default p
where not array_contains(
(select raw array_agg(distinct(p_id)) from default where p_id is not null)[0],id));
We get a list of all the ids and their children_count, zero or not. If you want more than just the id, add more fields or '*' to the select list for each query.