Mongoose Aggregate with Array of Objects in $lookup - node.js

Question

I have following Collections:
Collection A:
{
_id: 123,
name: "A",
amount: 2,
some_other_information: "ttt"
}
{
_id: 223,
name: "B",
amount: 2,
some_other_information: "ggg"
}
Collection B:
{
_id: 123,
name: "K",
amount: 2,
some_other_information: "fff"
}
{
_id: 2,
name: "L",
amount: 2,
some_other_information: "vvv"
}
Collection: D
{
_id: 123,
name: "test",
items: [
{
_id: 1,
a_id: 123,
b_id: 1,
c_id: 123
},
{
_id: 2,
a_id: 223,
b_id: 2,
c_id: 223
},
{
_id: 3,
a_id: 345,
b_id: 3
},
]
}
I want to aggregate Collection D with Collection A, with the a_id in the D collections in the items Array.
So that I have following output:
{
_id: 123,
name: "test",
items: [
{
_id: 1,
a: {
_id: 123,
name: "A",
amount: 2,
some_other_information: "ttt"
},
b: {
_id: 123,
name: "K",
amount: 2,
some_other_information: "fff"
},
c: {
_id: 123,
name: "A",
amount: 2,
some_other_information: "ttt"
}
},
{
_id: 2,
a: {
_id: 223,
name: "B",
amount: 2,
some_other_information: "ggg"
},
b: {
_id: 2,
name: "L",
amount: 2,
some_other_information: "vvv"
},
c: {
_id: 223,
name: "B",
amount: 2,
some_other_information: "ggg"
}
}
]
}
I tried
const x = await D.aggregate([
{
$lookup: {
from: "A",
localField: "items.a_id",
foreignField: "_id",
as: "a_items",
},
},
{
$project: {
_id: 1,
name: 1,
items: {
$map: {
input: "$a_items",
as: "ri",
in: {
$mergeObjects: [
"$$ri",
{
$arrayElemAt: [
{
$filter: {
input: "$items",
cond: {
$eq: [
"$$this._id",
"$$ri.a_id"
]
}
}
},
0
]
}
],
}
}
}
}
}]);
But in this way it is not aggregated with only the a_id. It shows every item in A Collection. Not to mention that it doesn't include c_id (which is the same as a_id, but it is not required, so it can be null).
So I don't know what to try anymore. Would be very helpful if someone can help.
Thanks in advance!

Answer 1

$lookup with collection A
$lookup with collection B
You can add more lookup if you want for collection C
$map to iterate loop of items array
show required fields in in
a add a field and filter items from a_items and get matching element and get first element
b add a field and filter items from b_items and get matching element and get first element
same as you can add c if you want
db.colD.aggregate([
{
$lookup: {
from: "colA",
localField: "items.a_id",
foreignField: "_id",
as: "a_items"
}
},
{
$lookup: {
from: "colB",
localField: "items.b_id",
foreignField: "_id",
as: "b_items"
}
},
{
$project: {
_id: 1,
name: 1,
items: {
$map: {
input: "$items",
as: "i",
in: {
_id: "$$i._id",
a: {
$arrayElemAt: [
{
$filter: {
input: "$a_items",
cond: { $eq: ["$$this._id", "$$i.a_id"] }
}
},
0
]
},
b: {
$arrayElemAt: [
{
$filter: {
input: "$b_items",
cond: { $eq: ["$$this._id", "$$i.b_id"] }
}
},
0
]
}
}
}
}
}
}
])
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