MongoDB lookup with object relation instead of array - mongodb

Question

I have a collection matches like this. I'm using players object {key: ObjectId, key: ObjectID} instead of classic array [ObjectId, ObjectID] for reference players collection
{
"_id": ObjectId("5eb93f8efd259cd7fbf49d55"),
"date": "01/01/2020",
"players": {
"home": ObjectId("5eb93f8efd259cd7fbf49d59"),
"away": ObjectId("5eb93f8efd259cd7fbf49d60")
}
},
{...}
And players collection:
{
"_id": ObjectId("5eb93f8efd259cd7fbf49d59"),
"name": "Roger Federer"
"country": "Suiza"
},
{
"_id": ObjectId("5eb93f8efd259cd7fbf49d60"),
"name": "Rafa Nadal"
"country": "España"
},
{...}
What's the better way to do mongoDB lookup? something like this is correct?
const rows = await db.collection('matches').aggregate([
{
$lookup: {
from: "players",
localField: "players.home",
foreignField: "_id",
as: "players.home"
}
},
{
$lookup: {
from: "players",
localField: "players.away",
foreignField: "_id",
as: "players.away"
},
{ $unwind: "$players.home" },
{ $unwind: "$players.away" },
}]).toArray()
I want output like this:
{
_id: 5eb93f8efd259cd7fbf49d55,
date: "12/05/20",
players: {
home: {
_id: 5eb93f8efd259cd7fbf49d59,
name: "Roger Federer",
country: "Suiza"
},
away: {
_id: 5eb93f8efd259cd7fbf49d60,
name: "Rafa Nadal",
country: "España"
}
}
}
{...}

Answer 1

You can try below aggregation query :
db.matches.aggregate([
{
$lookup: {
from: "players",
localField: "players.home",
foreignField: "_id",
as: "home"
}
},
{
$lookup: {
from: "players",
localField: "players.away",
foreignField: "_id",
as: "away"
}
},
/** Check output of lookup is not empty array `[]` & get first doc & write it to respective field, else write the same value as original */
{
$project: {
date: 1,
"players.home": { $cond: [ { $eq: [ "$home", [] ] }, "$players.home", { $arrayElemAt: [ "$home", 0 ] } ] },
"players.away": { $cond: [ { $eq: [ "$away", [] ] }, "$players.away", { $arrayElemAt: [ "$away", 0 ] } ] }
}
}
])
Test : mongoplayground
Changes or Issues with current Query :
1) As you're using two $unwind stages one after the other, If anyone of the field either home or away doesn't have a matching document in players collection then in the result you don't even get actual match document also, But why ? It's because if you do $unwind on [] (which is returned by lookup stage) then unwind will remove that parent document from result, To overcome this you need to use preservenullandemptyarrays option in unwind stage.
2) Ok, there is another way to do this without actually using $unwind. So do not use as: "players.home" or as: "players.away" cause you're actually writing back to original field, Just in case if you don't find a matching document an empty array [] will be written to actual fields either to "home" or "away" wherever there is not match (In this case you would loose actual ObjectId() value existing in that particular field in matches doc). So write output of lookup to a new field.
Or even more efficient way, instead of two $lookup stages (Cause each lookup has to go through docs of players collection again & again), you can try one lookup with multiple-join-conditions-with-lookup :
db.matches.aggregate([
{
$lookup: {
from: "players",
let: { home: "$players.home", away: "$players.away" },
pipeline: [
{
$match: { $expr: { $or: [ { $eq: [ "$_id", "$$home" ] }, { $eq: [ "$_id", "$$away" ] } ] } }
}
],
as: "data"
}
}
])
Test : mongoplayground
Note : Here all the matching docs from players which match with irrespective of away or home field will be pushed to data array. So to keep DB operation simple you can get that array from DB along with actual matches document & Offload some work to code which is to map respective objects from data array to players.home & players.away fields.