Find MongoDB documents that are not contained across arrays - mongodb

Question

MongoDB Collection A contains documents with an array with some document ids of collection B:
Collection A:
{
some_ids_of_b: ["id1", ...]
}
Collection B:
{
_id: "id1"
},
{
_id: "id2"
},
...
How do I query all documents from B whose _ids are NOT in contained in the some_ids_of_b arrays of documents of A?

Answer 1

Simple lookup from collection B to A and filter to keep only those documents where you don't find any matches.
db.collb.aggregate([
{
"$lookup": {
"from": "colla",
"localField": "_id",
"foreignField": "someIdsOfB",
"as": "a"
}
},
{
$match: {
$expr: {
$eq: [{$size: "$a"}, 0]
}
}
}
])
Demo

Answer 2

One option is:
db.collectionB.aggregate([
{$lookup: {
from: "collectionA",
let: {my_id: "$_id"},
pipeline: [
{$match: {$and: [
{_id: collADocId},
{$expr: {$in: ["$$my_id", "$some_ids_of_b"]}}
]}},
{$project: {_id: 1}}
],
as: "some_ids_of_b"
}},
{$match: {"some_ids_of_b.0": {$exists: false}}},
{$unset: "some_ids_of_b"}
])
See how it works on the playground example

Answer 3

You can do it with Aggregation Framework:
$group and $addToSet - To get all $some_ids_of_b from all the documents in A collection.
$set with $reduce - To create an array with all unique values of the IDs from the B collection.
$lookup - To fetch the documents from the B collection, where the _id of the document is not present in the $b_ids array.
$project - To project data as expected output.
db.A.aggregate([
{
"$group": {
"_id": null,
"b_ids": {
"$addToSet": "$some_ids_of_b"
}
}
},
{
"$set": {
b_ids: {
$reduce: {
input: "$b_ids",
initialValue: [],
in: {
$setUnion: [
"$$value",
"$$this"
]
}
}
}
}
},
{
"$lookup": {
from: "B",
let: {
b_ids: "$b_ids"
},
pipeline: [
{
"$match": {
"$expr": {
$ne: [
{
"$in": [
"$_id",
"$$b_ids"
]
},
true
]
}
}
}
],
as: "data"
}
},
{
"$project": {
data: 1,
_id: 0
}
}
])
Working Example