MongoDB: Populate reference in $group when aggregating? - mongodb

Question

I have a collection that I need to group by year. My aggregation pipeline is as such:
const WorkHistoryByYear = await this.workHistroyModel.aggregate([
{
$group: {
_id: '$year',
history: {
$push: '$$ROOT'
}
}
}
]);
Which works as expected, returning:
[{
"_id": 2003,
"history": [
{
"_id": "600331b3d84ac418877a0e5a",
"tasksPerformed": [
"5fffb180a477c4f78ad67331",
"5fffb18aa477c4f78ad67332"
],
"year": 2003
},
{
"_id": "600331dcd84ac418877a0e5c",
"tasksPerformed": [
"5fffb180a477c4f78ad67331"
],
"year": 2003
}
]
}]
but I'd like to populate a field if possible.
The WorkHistory schema has a field, tasksPerformed which is an array of ObjectId references. Here is the Task schema:
export const TaskSchema = new Schema({
active: {
type: Schema.Types.Boolean,
default: true,
},
title: {
type: Schema.Types.String,
required: true,
},
order: {
type: Schema.Types.Number,
index: true,
}
});
Is it possible to populate the referenced models within the aggregation? $lookup seems to be what I need, but I have yet to get that to work when following the documentation.
I don't do a lot of database work, so I'm having some difficulty finding the right operator(s) to use, and I've seen similar questions, but not a definitive answer.
Edit:
After adding the code from #varman's answer, my return is now:
{
"_id": 2003,
"history": {
"_id": "600331b3d84ac418877a0e5a",
"tasksPerformed": [
"5fffb180a477c4f78ad67331",
"5fffb18aa477c4f78ad67332"
],
"year": 2003,
"history": {
"tasksPerformed": []
}
}
}
I converted the ObjectId references to strings in an effort to help the matching, but I'm still coming up short.

Answer 1

You can do the lookup to join both collections
$unwind to deconstruct the array. (Array to Objects)
$lookup to join collections
$group to reconstruct the deconstructed array again
The script for the above result is
db.workHistory.aggregate([
{
"$unwind": "$history"
},
{
"$lookup": {
"from": "taskSchema",
"localField": "history.tasksPerformed",
"foreignField": "_id",
"as": "history.tasksPerformed"
}
},
{
"$group": {
"_id": "$_id",
"history": {
"$push": "$history"
}
}
}
])
Working Mongo playground
But before grouping, you have collection look like this
db={
"workHistory": [
{
"_id": 2003,
"history": [
{
"_id": "600331b3d84ac418877a0e5a",
"tasksPerformed": [
"5fffb180a477c4f78ad67331",
"5fffb18aa477c4f78ad67332"
],
"year": 2003
},
{
"_id": "600331dcd84ac418877a0e5c",
"tasksPerformed": [
"5fffb180a477c4f78ad67331"
],
"year": 2003
}
]
}
],
"taskSchema": [
{
"_id": "5fffb180a477c4f78ad67331",
"active": true,
"title": "first"
},
{
"_id": "5fffb18aa477c4f78ad67332",
"active": true,
"title": "second"
}
]
}
Since $unwind is expensive, we could have done aggregation
db.workHistory.aggregate([
{
"$lookup": {
"from": "taskSchema",
"localField": "tasksPerformed",
"foreignField": "_id",
"as": "tasksPerformed"
}
},
{
$group: {
_id: "$year",
history: {
$push: "$$ROOT"
}
}
}
])
Working Mongo playground

Answer 2

you could also do it without an unwind:
db.WorkHistory.aggregate(
[
{
$lookup: {
from: "Tasks",
localField: "tasksPerformed",
foreignField: "_id",
as: "tasksPerformed"
}
},
{
$group: {
_id: "$year",
history: { $push: "$$ROOT" }
}
}
])