How to use MongoDBs aggregate `$lookup` as `findOne()`

Question

So as you all know, find() returns an array of results, with findOne() returning just a simply object.

With Angular, this makes a huge difference. Instead of going {{myresult[0].name}}, I can simply just write {{myresult.name}}.

I have found that the $lookup method in the aggregate pipeline returns an array of results instead of just a single object.

For example, I have two colletions:

users collection:

[{
  "firstName": "John",
  "lastName": "Smith",
  "country": 123
}, {
  "firstName": "Luke",
  "lastName": "Jones",
  "country": 321
}]

countries collection:

[{
  "name": "Australia",
  "code": "AU",
  "_id": 123
}, {
  "name": "New Zealand",
  "code": "NZ",
  "_id": 321
}]

My aggregate $lookup:

db.users.aggregate([{
  $project: {
    "fullName": {
      $concat: ["$firstName", " ", "$lastName"]
    },
    "country": "$country"
  }
}, {
  $lookup: {
    from: "countries",
    localField: "country",
    foreignField: "_id",
    as: "country"
  }
}])

The results from the query:

[{
  "fullName": "John Smith",
  "country": [{
    "name": "Australia",
    "code": "AU",
    "_id": 123
  }]
}, {
 "fullName": "Luke Jones",
 "country": [{
   "name": "New Zealand",
   "code": "NZ",
   "_id": 321
 }]
}]

As you can see by the above results, each country is an array instead of a single object like "country": {....}.

How can I have my $lookup return a single object instead of an array since it will only ever match a single document?

Answer 1

You're almost there, you need to add another $project stage to your pipeline and use the $arrayElemAt to return the single element in the array.

db.users.aggregate(
    [
        {   "$project": {     
            "fullName": {       
                "$concat": [ "$firstName", " ", "$lastName"]     
            },
            "country": "$country"   
        }}, 
        { "$lookup": {     
                "from": "countries",     
                "localField": "country",     
                "foreignField": "_id",     
                "as": "countryInfo"   
        }}, 
        { "$project": { 
            "fullName": 1, 
            "country": 1, 
            "countryInfo": { "$arrayElemAt": [ "$countryInfo", 0 ] } 
        }} 
    ]
)

Answer 2

You can also use "preserveNullAndEmptyArrays"

Like so:

 db.users.aggregate(
        [
            {   "$project": {     
                "fullName": {       
                    "$concat": [ "$firstName", " ", "$lastName"]     
                },
                "country": "$country"   
            }}, 
            { "$lookup": {     
                    "from": "countries",     
                    "localField": "country",     
                    "foreignField": "_id",     
                    "as": "countryInfo"   
            }}, 
            {"$unwind": {
                    "path": "$countryInfo",
                    "preserveNullAndEmptyArrays": true
                }
            },
        ]
    )

Answer 3

db.users.aggregate([
    {
        $lookup: {
            from: 'countries',
            localField: 'country',
            foreignField: '_id',
            as: 'country'
        }
    },
    {
        $unwind: '$country'
    }
]).pretty()

You can use this mongo query for getting the country object

Answer 4

When you don't want to repeat all fields in project, just overwrite the field in question with $addFields:

db.users.aggregate([
    {   "$project": {     
        "fullName": {       
            "$concat": [ "$firstName", " ", "$lastName"]     
        },
        "country": "$country"   
    }}, 
    { "$lookup": {     
        "from": "countries",     
        "localField": "country",     
        "foreignField": "_id",     
        "as": "countryInfo"   
    }},
    { "$addFields": {
        "countryInfo": {
            "$arrayElemAt": [ "$countryInfo", 0 ]
        }
    }}
])