MongoDB查询联接两个集合

Question

I´m trying to make a query to my bbdd in order to get some info that involve two collections. First, I have one collecction, called Collectables that is a collection that store all available items that One user can get using an App. For example, this collection can have... 100 items. This is the maximum number of items. This is one document of this collection (called collectables)

{
"_id" : ObjectId("5d387ecfbb676b173aa57fe3"),
"img" : "some url",
"name" : "La 5",
"__v" : 0,
"amount" : 17,
"available" : 16,
"collec" : ObjectId("5d36c0c34c86991db93bd7c8"),
"gen" : 3,
"metadata" : {},
"position" : 1
}

Then, I have another collection called AppUsers. In this collection I store all info related to the user. Each user of the App has his own record here. The point is that besides meta info of the user such alias, avatar, age... I have one field called collectables. Is an array. Here I store what collectable have each user. For example, if one user have 10 collectables (from the other collection) I have 10 entries in this array with that info. Is possible that one user "win" the same item (collectable) twice or more.. so in this collection I store a count field with the total. For example, a user win the collectable with Id 1 the first time, so I add the entry in the array with count 1. If then the user win the same collectable, the count is 2... and so on.

This is an example of one user... with 3 collectables, but several number of each item.

{
"_id" : ObjectId("5d36dc9445526a215c4eff52"),
"twitter" : "1",
"alias" : "ViktorCrowley",
"__v" : 25,
"collectables" : [ 
    {
        "count" : 12,
        "collectable" : ObjectId("5d36c1ba4c86991db93bd7e7")
    }, 
    {
        "count" : 25,
        "collectable" : ObjectId("5d36c13d4c86991db93bd7c9")
    }, 
    {
        "count" : 8,
        "collectable" : ObjectId("5d381e122f25221126a98f9c")
    }
   ]
}

So, in this case, this user, for example have 3 differents items (collectables). But imagine that the total of collectables from the first collection is 100.

Now... what I´m looking for. I need a query that give me (paginated) the items from the first collection (collectables) and in the case that the user already have one of this items, marked with the total count. I mean, I want all the items from the first collection, with a new field, called count. If the user doens´t have any entry in his array, count will be 0, and if the user for that item, has for example 4 collectables, the count will be 4. Some thing like this:

[
{
"_id" : ObjectId("5d387ecfbb676b173aa57fe3"),
"img" : "some url",
"name" : "THe one",
"__v" : 0,
"amount" : 17,
"available" : 16,
"collec" : ObjectId("5d36c0c34c86991db93bd7c8"),
"gen" : 3,
"metadata" : {},
"position" : 1,
"count" : 0
},
{
"_id" : ObjectId("5d387ecfbb676b173aa57fe3"),
"img" : "some url",
"name" : "The two",
"__v" : 0,
"amount" : 17,
"available" : 16,
"collec" : ObjectId("5d36c0c34c86991db93bd7c8"),
"gen" : 3,
"metadata" : {},
"position" : 2,
"count" : 1
},
{
"_id" : ObjectId("5d387ecfbb676b173aa57fe4"),
"img" : "some url",
"name" : "The Three",
"__v" : 0,
"amount" : 17,
"available" : 16,
"collec" : ObjectId("5d36c0c34c86991db93bd7c8"),
"gen" : 3,
"metadata" : {},
"position" : 3,
"count" : 0
},
{
"_id" : ObjectId("5d387ecfbb676b173aa57fe4"),
"img" : "Some url",
"name" : "La 5",
"__v" : 0,
"amount" : 17,
"available" : 16,
"collec" : ObjectId("5d36c0c34c86991db93bd7c8"),
"gen" : 3,
"metadata" : {},
"position" : 4,
"count" : 12
}

I tried several things using aggregate and lookup but I can´t get make it work.

The only I could get, was retrieve the info from AppUser and the total count..

Something like this (with mongoose):

AppUser.aggregate([
    { $match: matchQuery },
    {$unwind: "$collectables"},

    {
        $lookup: 
        {
            from: "collectables",
            localField: "collectables.collectable",
            foreignField: "_id",
            as: "result"
        }
    },
    { $sort: { "result.position": 1 } },
    {$unwind: "$result"},
    { $addFields : { "result.count" : "$collectables.count" } 
},
    { $replaceRoot: { newRoot: "$result" } },
    { $skip: size * (page - 1) },
    { $limit: size }
 ]).exec((err, result) => 
{
    if (err) 
    {
        console.log(err);
        return res.status(401).send({ success: false });
    }
    else {
        return res.status(200).send({ success: true, result });
    }

});

So I need help because i have three days with this and I can´t get something nice... Thanks in advance...

Answer 1

Below query will bring us the expected output

db.appuser.aggregate([
  { $unwind: "$collectables" },
  {
    $lookup: {
      from: "collectables",
      localField: "collectables.collectable",
      foreignField: "_id",
      as: "result"
    }
  },
  {
    $project: { 
      result: 1,
      flag: { "$gt": [ {"$size": "$result"}, 0 ] },
      "collectables.count": 1,
      alias: 1,
      _id: 0
    }
  },
  {
    $match: { flag: true }
  }
]);

Unwind the appuser.collectables array, then do a lookup between the collections collectables and appuser , then use $project and $match to get the desired result.

In the $project stage added a flag to filter out the desired documents, by checking the size of the result array of the $lookup output.

Sample Data used

db.collectables.find()


{
  "_id": "5d387ecfbb676b173aa57fe3",
  "img": "some url",
  "name": "La 5",
  "__v": 0,
  "amount": 17,
  "available": 16,
  "collec": "5d36c0c34c86991db93bd7c8",
  "gen": 3,
  "metadata": {

  },
  "position": 1
}

db.appuser.find()

[{
  "_id": "5d36dc9445526a215c4eff52",
  "twitter": "1",
  "alias": "ViktorCrowley",
  "__v": 25,
  "collectables": [
    {
      "count": 12,
      "collectable": "5d36c1ba4c86991db93bd7e7"
    },
    {
      "count": 25,
      "collectable": "5d36c13d4c86991db93bd7c9"
    },
    {
      "count": 8,
      "collectable": "5d381e122f25221126a98f9c"
    }
  ]
},
{
  "_id": "5d36dc9445526a215c4efga1",
  "twitter": "1",
  "alias": "John",
  "__v": 11,
  "collectables": [
    {
      "count": 10,
      "collectable": "5d387ecfbb676b173aa57fe3"
    }
  ]
},
{
  "_id": "6dc2dc9445526a215c4efga1",
  "twitter": "1",
  "alias": "Alice",
  "__v": 11,
  "collectables": [
    {
      "count": 25,
      "collectable": "5d387ecfbb676b173aa57fe3"
    },
    {
      "count": 3,
      "collectable": "5d36c13d4c86991db9312349"
    },
    {
      "count": 5,
      "collectable": "5d381e122f25221126a9711c"
    }
  ]
}]

Final Output

{
  "alias": "John",
  "collectables": {
    "count": 10
  },
  "result": [
    {
      "_id": "5d387ecfbb676b173aa57fe3",
      "img": "some url",
      "name": "La 5",
      "__v": 0,
      "amount": 17,
      "available": 16,
      "collec": "5d36c0c34c86991db93bd7c8",
      "gen": 3,
      "metadata": {

      },
      "position": 1
    }
  ],
  "flag": true
}
{
  "alias": "Alice",
  "collectables": {
    "count": 25
  },
  "result": [
    {
      "_id": "5d387ecfbb676b173aa57fe3",
      "img": "some url",
      "name": "La 5",
      "__v": 0,
      "amount": 17,
      "available": 16,
      "collec": "5d36c0c34c86991db93bd7c8",
      "gen": 3,
      "metadata": {

      },
      "position": 1
    }
  ],
  "flag": true
}

Hope it helps!