Use Result From Previous Query In Another Query In Mongodb

1st query: db.a.find({ 'type': ObjectId('50ed90f5a70defef23000002'), 'config.name': 'alpha' }, { 'user': 1 }) 1st query result: { '_id' : ObjectId('5684035c91b46

Solution 1:

You can use an aggregation with a $match to match your condition and $lookup to map your local field user to your usercollection _id field :

db.a.aggregate(
    [{
        $match: {
            "type": ObjectId("50ed90f5a70defef23000002"),
            "config.name": "alpha"
        }
    }, {
        $lookup: {
            from: "users",
            localField: "user",
            foreignField: "_id",
            as: "users"
        }
    }, {
        $unwind: "$users"
    }]);

In Javascript, with mongoose for example you can do this with :

YourModel.aggregate(
    [{
        $match: {
            "type": ObjectId("50ed90f5a70defef23000002"),
            "config.name": "alpha"
        }
    }, {
        $lookup: {
            from: "users",
            localField: "user",
            foreignField: "_id",
            as: "users"
        }
    }, {
        $unwind: "$users"
    }],
    function(err, result) {
        console.log("lastname : " + result.users.lastname);
    });

Solution 2:

Since MongoDB doesn't support joins, this can't be conbined into one query.

You can do this directly from the mongo shell very easily by firing first query, storing the result in a variable and then using it to fire second query.

var userDoc = db.a.findOne({"type" : ObjectId("50ed90f5a70defef23000002"),"config.name":"alpha" },{"user":1});
db.users.find({"_id" : userDoc.user},{"lastname":1})

If your first query returns multiple results, then

var cursor = db.a.find({"type" : ObjectId("50ed90f5a70defef23000002"),"config.name":"alpha" },{"user":1});
  cursor.forEach(function (userDoc) {
  var user = db.users.findOne({"_id" : userDoc.user},{"lastname":1});
  printjson(user);
})

Note: MongoDB supports left outer join using $lookup in aggregation pipeline since version 3.2

but in this case, that seems to be an overkill.