MongoDB select distinct where not in select distinct - mongodb

Question

In a mongoDB bd, I need to find all the records where those records aren't in a different collection
Say I have 2 collections
1) user_autos
{
make: string,
user_id: objId
}
2) auto_makes
{
mfg: string,
make: string
}
I need to find all the "makes" that are not part of the "master makes" list
I want to do the parallel to this SQL
SELECT DISTINCT
a.make
FROM
user_autos a
WHERE
a.make NOT IN (
SELECT DISTINCT
b.make
FROM
auto_makes b
)
Help please

Answer 1

to achieve this, you need to make use of aggragation with pipeline stage 'lookup'.
lookup does left join between two collections. so, obviously the unmatching documents of
'user_autos' gives an empty nested array 'auto_makes'. and then 'group' the 'user_autos'
with 'make'. so that a list of 'user_auto' documents will be resulted.
you can do it as below.
db.user_autos.aggregate([
{$lookup:{
from:"äuto_makes",
localField:"make",
foreignField:"make",
as:"m"
}},
{$match:{
m:{$exists:false}
}},
{$group:{
_id:"$make"
}}
//if you want to get the distinct 'make' values as an array of single
//document, add another $group stage.
{$group:{
_id:"",
make_list:{$addToSet:"$_id"}
}}
])
Visit https://docs.mongodb.com/manual/reference/operator/aggregation/group/ ,
https://docs.mongodb.com/manual/reference/operator/aggregation/lookup/