Mongo db not in query by having two subset of documents from same collection - mongodb

Question

I am new to mongodb. Assume the following. There are 3 types of documents in one collection x, y and z.
docs = [{
"item_id": 1
"type": "x"
},
{
"item_id": 2
"type": "x"
},{
"item_id": 3
"type": "y",
"relavent_item_ids": [1, 2]
},
{
"item_id": 3
"type": "y",
"relavent_item_ids": [1, 2, 3]
},{
"item_id": 4
"type": "z",
}]
I want to get the following.
Ignore the documents with type z
Get all the documents of type x where it's item_id is not in relavent_item_ids of type y documents.
The result should have item_id field.
I tried doing match $in but this returns me all the records, I am unable to figure out how to have in condition with subset of documents of type y.

Answer 1

You can use below query
const item_ids = (await db.collection.find({ "type": "y" })).map(({ relavent_item_ids }) => relavent_item_ids)
const result = db.collection.find({
"item_id": { "$exists": true },
"type": { "$ne": "z", "$eq": "x" },
"relavent_item_ids": { "$nin": item_ids }
})
console.log({ result })
Ignore the documents with type z --> Use $ne not equal to query operator to filter out z types.
Get all the documents of type x where it's item_id is not in relavent_item_ids of type y documents --> Use $expr to match the same documents fields.
The result should have item_id field --> Use $exists query operator.

Answer 2

The solution:
db.test.aggregate( [
{
$facet: {
firstQuery: [
{
$match: { type: { $eq: "x", $ne: "z" } }
},
{
$project: {
item_id : 1, _id: 0
}
}
],
secondQuery: [
{
$match: { type: "y" }
},
{
$group: {
_id: null,
relavent: { $push: "$relavent_item_ids" }
}
},
{
$project: {
relavent: {
$reduce: {
input: "$relavent",
initialValue: [ ],
in: { $setUnion: [ "$$value", "$$this" ] }
}
}
}
}
]
}
},
{
$addFields: { secondQuery: { $arrayElemAt: [ "$secondQuery", 0 ] } }
},
{
$project: {
result: {
$filter: {
input: "$firstQuery" ,
as: "e",
cond: { $not: [ { $in: [ "$$e.item_id", "$secondQuery.relavent" ] } ] }
}
}
}
},
] )
Using the input documents in the question post and adding one more following document to the collection:
{
"item_id": 11,
"type": "x",
}
: only this document's item_id (value 11) will show in the output.
The aggregation uses a $facet to make two individual queries with a single pass. The first query gets all the "x" types (and ignores type "z") as an array. The second query gets an array of relavent_item_ids with unique values (from the documents of type "y"). The final, $project stage filters the first query result array with the condition:
Get all the documents of type x where it's item_id is not in
relavent_item_ids of type y documents

Answer 3

I am not sure if its an elegant solution.
db.getCollection('test').aggregate([
{
"$unwind": {
"path": "$relavent_item_ids",
"preserveNullAndEmptyArrays": true
}
},
{
"$group": {
"_id":null,
"relavent_item_ids": {"$addToSet":"$relavent_item_ids"},
"other_ids": {
"$addToSet":{
"$cond":[
{"$eq":["$type", "x"]},
"$item_id",
null
]
}
}
}
},
{
"$project":{
"includeIds": {"$setDifference":["$other_ids", "$relavent_item_ids"]}
}
},
{
"$unwind": "$includeIds"
},
{
"$match": {"includeIds":{"$ne":null}}
},
{
"$lookup":{
"from": "test",
"let": { "includeIds": "$includeIds"},
"pipeline": [
{ "$match":
{ "$expr":
{ "$and":
[
{ "$eq": [ "$item_id", "$$includeIds" ] },
{ "$eq": [ "$type", "x" ] }
]
}
}
}
],
"as": "result"
}
},
{
"$unwind": "$result"
},
])