如何使用聚合从 mongoDB 中对象的嵌套数组中提取信息

Question

I am trying to get the hang of the aggregation in mongoDB, I am new at this and I think I kinda got lost. I have a collection of users, each user has an array of social networks, and each item in this array is an object. How do I get value only from a given social network/s object/s.

{
 "user":{
     "_id": "d10430c8-e59",
     "username": "John",
     "password": "f7wei93",
     "location": "UK",
     "gender": "Male",
     "age": 26,
     "socials": [
        {
            "type": "instagram",
            "maleFollowers": 23000,
            "femaleFollowers": 65000,
            "posts": 5400,
            "avgFollowerAge": 22
        },
        {
            "type": "facebook",
            "maleFollowers": 4000,
            "femaleFollowers": 6700,
            "posts": 330,
            "avgFollowerAge": 25
        },
        {
            "type": "snapchat",
            "maleFollowers": 873,
            "femaleFollowers": 1200,
            "posts": 1200,
            "avgFollowerAge": 21
        },

     ]
 }
}

I want to get the totalFollowersCount ( maleFollowers + femaleFollowers) for each social network type that is given in an array. for example ["instagram", "snapchat"] will only give me the totalFollowersCount for this two specific networks.

I started playing around with aggregation but didn't figure it out all the way.

Answer 1

First, we need to flatten the socials array. Then, we group by socials type + 'user' fields and sum( maleFollowers + femaleFollowers) .

With $match operator, we filter desired social networks.

db.users.aggregate([
  {
    $unwind: "$user.socials"
  },
  {
    $group: {
      _id: {
        user: "$user._id",
        type: "$user.socials.type"
      },
      "totalFollowersCount": {
        $sum: {
          $add: [
            "$user.socials.femaleFollowers",
            "$user.socials.maleFollowers"
          ]
        }
      }
    }
  },
  {
    $match: {
      "_id.type": {
        $in: [
          "instagram",
          "snapchat"
        ]
      }
    }
  }
])

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