MongoDB $lookup in different collections by _id - node.js

Question

I have 3 mongoDB collections
I need to aggregate them with $lookup operator but I didn't find anything/**or I'm bad looking **
1st one is suppliers
{
"_id" : ObjectId("111"), //for example, in db is mongodb ids
"name" : "supplier 1",
}
{
"_id" : ObjectId("222"),
"name" : "supplier 1",
}
2nd one is clients
{
"_id" : ObjectId("333"), //for example, in db is mongodb ids
"name" : "clients 1",
}
{
"_id" : ObjectId("444"),
"name" : "clients 2",
}
and 3rd is moves
{
"_id" : ObjectId("..."), //for example, in db is mongodb ids
"moveName" : "move 1",
"agent": ObjectId("111") // this is from suppliers collection
}
{
"_id" : ObjectId("..."),
"moveName" : "move 2",
"agent": ObjectId("333") // this one is from CLIENTS collection
}
so like output I need data like this
{
"_id" : ObjectId("..."), //for example, in db is mongodb ids
"moveName" : "move 1",
**"agent": supplier 1** // this is from suppliers collection
}
{
"_id" : ObjectId("..."),
"moveName" : "move 2",
**"agent": clients 1** // this one is from CLIENTS collection
}
back end is nodejs, I`m using mongoose, how I can search in 2nd collection if noresult in 1st?
const moves = await Move.aggregate([
{ $match: query }, // here all wokrs good
{
$lookup: {
from: 'clients',
localField: 'agent',
foreignField: '_id',
as: 'agent'
}
},{ $unwind: {path: "$agent" , preserveNullAndEmptyArrays: true} },
{
$lookup: {
from: 'suppliers',
localField: 'agent',
foreignField: '_id',
as: 'agent2'
}
},
{
$project: {
operationName: 1,
agent: {$ifNull: ['$agent.name', '$agent2.name']}
}
}
])
Thank You!

Answer 1

As suggested by #hhharsha36, we can use $facet operator which allows to run several pipelines within a single stage.
Explanation
facet
suppliers = $lookup suppliers collection and filter only matched results
clientes = $lookup clientes collection and filter only matched results
concatArrays = We concat suppliers and clients results into a single movies array
unwind = We flatten movies array [a, b, c] -> a
b
c
replaceWith = We replace the root element [movies:a, movies:b -> a, b]
mergeObject = allows us to pick the agent name (this way we avoid 1 more stage)
db.moves.aggregate([
{
$facet: {
suppliers: [
{
$lookup: {
from: "suppliers",
localField: "agent",
foreignField: "_id",
as: "agent"
}
},
{
$match: {
agent: {
$not: {
$size: 0
}
}
}
}
],
clients: [
{
$lookup: {
from: "clients",
localField: "agent",
foreignField: "_id",
as: "agent"
}
},
{
$match: {
agent: {
$not: {
$size: 0
}
}
}
}
]
}
},
{
$project: {
movies: {
"$concatArrays": [
"$clients",
"$suppliers"
]
}
}
},
{
$unwind: "$movies"
},
{
$replaceWith: {
"$mergeObjects": [
"$movies",
{
agent: {
"$arrayElemAt": [
"$movies.agent.name",
0
]
}
}
]
}
}
])
MongoPlayground

Answer 2

This aggregation query gives the desired result:
db.moves.aggregate([
{
$lookup: {
from: "suppliers",
localField: "agent",
foreignField: "_id",
as: "moves_sup"
}
},
{
$unwind: { path: "$moves_sup" , preserveNullAndEmptyArrays: true }
},
{
$lookup: {
from: "clients",
localField: "agent",
foreignField: "_id",
as: "moves_client"
}
},
{
$unwind: { path: "$moves_client" , preserveNullAndEmptyArrays: true }
},
{
$addFields: {
agent: {
$cond: [ { $eq: [ { $type: "$moves_sup" }, "object" ] },
"$moves_sup.name",
{ $cond: [ { $eq: [ { $type: "$moves_client" }, "object" ] }, "$moves_client.name", "undefined" ] }
] },
moves_client: "$$REMOVE",
moves_sup: "$$REMOVE"
}
},
])