How to lookup inside lookup in MongoDB Aggregate? - node.js

Question

I have a simple 3 collections. This bellow is their pseudocode. I want to get all shipments and for each shipment, I want to have all bids for that shipment and in each bid, I need userDetails object.
User: {
name: string,
}
Shipment: {
from: string,
to: string
}
Bid: {
amount: number,
shipmentId: Ref_to_Shipment
userId: Ref_to_User
}
This is what I have tried:
const shipments = await ShipmentModel.aggregate([
{
$lookup: {
from: "bids",
localField: "_id",
foreignField: "shipmentId",
as: "bids"
}
},
{
$lookup: {
from: "users",
localField: "bids.userId",
foreignField: "_id",
as: "bids.user"
}
}
])
And I got the following result:
[
{
"_id": "5fad2fc04458ac156531d1b1",
"from": "Belgrade",
"to": "London",
"__v": 0,
"bids": {
"user": [
{
"_id": "5fad2cdb4d19c80d1b6abcb7",
"name": "Amel",
"email": "Muminovic",
"password": "d2d2d2",
"__v": 0
}
]
}
}
]
I am trying to get all Shipments with their bids and users within bids. Data should look like:
[
{
"_id": "5fad2fc04458ac156531d1b1",
"from": "Belgrade",
"to": "London",
"__v": 0,
"bids": [
{
"_id": "5fad341887c2ae1feff73402",
"amount": 400,
"userId": "5fad2cdb4d19c80d1b6abcb7",
"shipmentId": "5fad2fc04458ac156531d1b1",
"user": {
"name": "Amel",
}
"__v": 0
}
]
}
]

Answer 1

Try with the following code:
const shipments = await ShipmentModel.aggregate([
{
$lookup: {
from: "bids",
localField: "_id",
foreignField: "shipmentId",
as: "bids"
}
},
{
$unwind: {
path: "$bids",
preserveNullAndEmptyArrays: true
}
},
{
$lookup: {
from: "users",
localField: "bids.userId",
foreignField: "_id",
as: "bids.user"
}
}
])
If you want to prevent null and empty arrays then set
preserveNullAndEmptyArrays: false

Answer 2

Try this query and chek if works and the behaviour is as you expected:
db.Shipment.aggregate([
{
$lookup: {
from: "Bid",
localField: "id",
foreignField: "shipmentId",
as: "bids"
}
},
{
$lookup: {
from: "user",
localField: "id",
foreignField: "id",
as: "newBids"
}
},
{
$project: {
"newBids.id": 0,
"newBids._id": 0,
}
},
{
$match: {
"bids.userId": 1
}
},
{
$addFields: {
"newBids": {
"$arrayElemAt": [
"$newBids",
0
]
}
}
},
{
$set: {
"bids.user": "$newBids"
}
},
{
$project: {
"newBids": 0
}
}
])
This query do your double $lookup and then a $project to delete the fields you don't want, and look for the userId to add the field user. As $lookup generate an array, is necessary use arrayElemAt to get the first position.
Then $set this value generated into the object as bids.user and remove the old value.
Note that I have used a new field id instead of _id to read easier the data.

Answer 3

Try this
I figured out it based on MongoDB $lookup on array of objects with reference objectId and in the answer from J.F. (data organization). Note that he used id instead of _id
The code is
db.Shipment.aggregate([
{
$lookup: {
from: "Bid",
localField: "id",
foreignField: "shipmentId",
as: "bids"
}
},
{
$lookup: {
from: "user",
localField: "bids.userId",
foreignField: "id",
as: "allUsers"
}
},
{
$set: {
"bids": {
$map: {
input: "$bids",
in: {
$mergeObjects: [
"$$this",
{
user: {
$arrayElemAt: [
"$allUsers",
{
$indexOfArray: [
"$allUsers.id",
"$$this.userId"
]
}
]
}
}
]
}
}
}
}
},
{
$unset: [
"allUsers"
]
},
// to get just one
//{
// $match: {
// "id": 1
// }
// },
])