How to use aggregate map and filter for multiple arrays in Mongodb

Question

Here i have Two arrays,

 Usercar = [{
            parentId :001
            cars:[
                   {_id: 1, name: bmw, color: red},
                   {_id: 2, name: Ford, color: black},
                   {_id: 3, name: Volkswagen, color: black},
                 ]
              }]


 Userfavorite = 
               [{
                  parentId :001,
                  favoriteCars:[1,3] //mongoose.Types.ObjectId
               }]

I want o show user favorite cars using mongodb aggregate, here is my code

let carsId= [1,3];


    {$match: {
        parentId :001
    }},
    {
        $project:{
            cars:{
                $filter:{

                        input:"$cars",
                        as :'cars',
                        cond:{ $eq :["$$cars._id", mongoose.Types.ObjectId('1')]}
                      //cond:{ $eq :["$$cars._id", carsId]}
                }
            }
        }
    }

the above code only work, when pass single carsId, I want user favorite Cars details from Usercar's collection, how to do that in mongodb aggregate ?

Are Usercar and Userfavorite in the same document, or separate collections? — Joe
– Joe, Commented Apr 22, 2020 at 5:24

Joe · Accepted Answer · 2020-04-22 06:07:51Z

1

One possible aggregate that could do this is:

aggregate on the Usercar collection
lookup the matching document from the Userfavorite collection
in case there are multiple favorite documents, combine them into a single favorite array
reduce the cars collection to find those matching the favorites, and add to an array

db.Usercar.aggregate([
  {$lookup: {
      from: "Userfavorite",
      localField: "parentId",
      foreignField: "parentId",
      as: "favorite"
  }},
  {$set: {
      favorite: {
        $arrayElemAt: [
          {$concatArrays: "$favorite.favoriteCars"},
          0
        ]
      }
  }},
  {$addFields: {
      favoriteCarDetails: {
        $filter: {
          input: "$cars",
          cond: {
            $in: [
              "$$this._id",
              "$favorite"
            ]
          }
        }
      }
  }}
])

Playground

edited Apr 22, 2020 at 6:07

answered Apr 22, 2020 at 5:41

Joe

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3 Comments

Harry Edward Over a year ago

favoriteCarDetails is empty :(

Harry Edward Over a year ago

why we need concatArrays in the case ?

Joe Over a year ago

You don't necessarily, that just deals with the possibility that there are 2 matching documents in the other collection. If there will only ever be one, you can just project it or unwind it to pull it out of the looked up array

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