Mongo how to $lookup with DBRef

Question

I have a trouble(/(ㄒoㄒ)/~~). Suppose that collection A is

{ 
    \"_id\" : ObjectId(\"582abcd85d2dfa67f44127e1\"), 
    \"bid\" : [
        DBRef(\"B\",         


        

Answer 1

Actually, the other answer is wrong. It is possible to do a lookup on a DBref field within your aggregator, and you don't need mapreduce for that.

Solution

db.A.aggregate([
{
    $project: { 
        B_fk: {
          $map: { 
             input: { 
                  $map: {
                      input:"$bid",
                      in: {
                           $arrayElemAt: [{$objectToArray: "$$this"}, 1]
                      },
                  }
             },
             in: "$$this.v"}},
        }
}, 
{
    $lookup: {
        from:"B", 
        localField:"B_fk",
        foreignField:"_id", 
        as:"B"
    }
}
])

result

{
    "_id" : ObjectId("59bb79df1e9c00162566f581"),
    "B_fk" : null,
    "B" : [ ]
},
{
    "_id" : ObjectId("582abcd85d2dfa67f44127e1"),
    "B_fk" : [
        ObjectId("582abcd85d2dfa67f44127e0"),
        ObjectId("582abcd85d2dfa67f44127e1")
    ],
    "B" : [
        {
            "_id" : ObjectId("582abcd85d2dfa67f44127e0"),
            "status" : NumberInt("1"),
            "seq" : NumberInt("0")
        }
    ]
}

Short Explanation

Loop through the DBRefs with $map, break each DBref into an array, keep only the $id field, then get rid of the k:v format with $$this.v, keeping only the ObjectId and removing all the rest. You can now lookup on the ObjectId.

Step-by-step Explanation

Within the aggregator, a DBRef BSON type can be handled like an object, with two or three fields (ref, id, and db).

If you do:

db.A.aggregate([
    {
        $project: { 
            First_DBref_as_array: {$objectToArray:{$arrayElemAt:["$bid",0]}},
            Second_DBref_as_array: {$objectToArray:{$arrayElemAt:["$bid",1]}},
            }

    },

])

This is the result:

{
"_id" : ObjectId("582abcd85d2dfa67f44127e1"),
"First_DBref_as_array : [
    {
        "k" : "$ref",
        "v" : "B"
    },
    {
        "k" : "$id",
        "v" : ObjectId("582abcd85d2dfa67f44127e0")
    }
],
"Second_DBref_as_array" : [
    {
        "k" : "$ref",
        "v" : "B"
    },
    {
        "k" : "$id",
        "v" : ObjectId("582abcd85d2dfa67f44127e0")
    }
]
}

Once you have transformed a dbref into an array, you can get rid of the useless fields by querying only the value at index 1, like this:

db.A.aggregate([
    {
        $project: { 
            First_DBref_as_array: {$arrayElemAt: [{$objectToArray:{$arrayElemAt:["$bid",0]}},1]},
            Second_DBref_as_array: {$arrayElemAt: [{$objectToArray:{$arrayElemAt:["$bid",0]}},1]},
            }

    },

])

result:

{
    "_id" : ObjectId("582abcd85d2dfa67f44127e1"),
    "First_DBref_as_array" : {
        "k" : "$id",
        "v" : ObjectId("582abcd85d2dfa67f44127e0")
    },
    "Second_DBref_as_array" : {
        "k" : "$id",
        "v" : ObjectId("582abcd85d2dfa67f44127e0")
    }
}

Then you can get finally to the value you want by pointing to "$myvalue.v", just like this

db.A.aggregate([
    {
        $project: { 
            first_DBref_as_array: {$arrayElemAt: [{$objectToArray:{$arrayElemAt:["$bid",0]}},1]},
            second_DBref_as_array: {$arrayElemAt: [{$objectToArray:{$arrayElemAt:["$bid",0]}},1]},
            }

    },
    {
        $project: {
            first_DBref_as_ObjectId: "$first_DBref_as_array.v",
            second_DBref_as_ObjectId: "$second_DBref_as_array.v"
        }
    }

])

result:

{
    "_id" : ObjectId("582abcd85d2dfa67f44127e1"),
    "first_DBref_as_ObjectId" : ObjectId("582abcd85d2dfa67f44127e0"),
    "second_DBref_as_ObjectId" : ObjectId("582abcd85d2dfa67f44127e0")
}

Obviously, in a normal pipeline, you don't need all these redundant steps, using a nested $map, you can get to the same result in one go :

db.A.aggregate([
    {
        $project: { 
            B_fk: { $map : {input: { $map: {    input:"$bid",
                                    in: { $arrayElemAt: [{$objectToArray: "$$this"}, 1 ]}, } },
                            in: "$$this.v"}},

            }
    }, 

])

result:

{
    "_id" : ObjectId("582abcd85d2dfa67f44127e1"),
    "B_fk" : [
        ObjectId("582abcd85d2dfa67f44127e0"),
        ObjectId("582abcd85d2dfa67f44127e1")
    ]
}

I hope the explanation is clear enough, if not feel free to ask.

Answer 2

As of mongoDB 3.4, this is not possible. You can't use DBRef in the aggregation pipeline, except in the $match stage.

I strongly recommend you to get rid of DBRef and switch to manual references. However, if you really need to keep DBRef, here is an (ugly) solution:

first, create a new collection named "C", where DBRefs are replaced by their Ids using mapReduce:

db.A.mapReduce(
    function() {
        var key = this._id; 
        var value = [];  
        for ( var index = 0; index < this.bid.length; index++){
           value.push(this.bid[index].$id); 
        }
        emit(key, value); 
    },
    function(key,values) {
        return  values;
    },
    {
        "query": {},
        "out": "C" 
    }
)

then, run your aggregation query on the new "C" collection:

db.C.aggregate([
   {
      $unwind:"$value"
   },
   {
      $lookup:{
         from:"B",
         localField:"value",
         foreignField:"_id",
         as:"bs"
      }
   }
]);

output:

    {
       "_id":ObjectId("582abcd85d2dfa67f44127e1"),
       "value":ObjectId("582abcd85d2dfa67f44127e0"),
       "bs":[
          {
             "_id":ObjectId("582abcd85d2dfa67f44127e0"),
             "status":1,
             "seq":0
          }
       ]
    }{
       "_id":ObjectId("582abcd85d2dfa67f44127e1"),
       "value":ObjectId("582abcd85d2dfa67f44127e1"),
       "bs":[
          {
             "_id":ObjectId("582abcd85d2dfa67f44127e1"),
             "status":1,
             "seq":0
          }
       ]
    }