MongoDB aggregate project return an array of _id [duplicate]

Question

As we know if we want to get an array of _id we can do:

db.collections.distinct("_id");

My question is how can I get an array of _id if I need to do a complicate logic with aggregate. Ex:

db.getCollection('users').aggregate({
        $match : {
            is_register_completed : { $ne : true}
        }
    }
    //other operator like $lookup, $group
    ,
     {
         $project : {_id:1}
    }
     )

I get

{
"_id" : "1",
"_id" : "2"
}

what i want is just like we do distinct

{[1,2]}

Updated: this is what i try to do with $group

db.getCollection('users').aggregate({
        $match : {
            is_register_completed : { $ne : true}
        }
    },
    {
        $group: {
        _id:null, all:{$addToSet: "$_id"}
        }
     }, {$project: {_id:0,all:1}}
     )

but i still get

{
all : ["1","2"]
}

or I can do .map(function(el) { return el._id }) after getting

{
    "_id" : "1",
    "_id" : "2"
    }

, but the map is client side transformation which I think it will affect the performance.

Can you please put an "edit" or "update" tag in your question for the newly updated code? My answer might get downvoted under the impression that its already included in your question. — Dushyant Bangal
– Dushyant Bangal, Commented Oct 6, 2017 at 6:21
With distinct() you just get an array of elements [1, 2] for example, not {[1, 2]} as you implied which is obviously invalid JSON. You can manipulate the result you get from the aggregate operation to return the array. — chridam
– chridam, Commented Oct 6, 2017 at 6:22
@chridam yes [1,2] is what i want, like I mention I would not prefer to manipulate the result using map in client side. — Chi Dov
– Chi Dov, Commented Oct 6, 2017 at 6:26
Do as much as you can on the mongodb. Why make the node server busy when you can make the async call to mongodb do the work? — Dushyant Bangal
– Dushyant Bangal, Commented Oct 6, 2017 at 6:42

Dushyant Bangal · Accepted Answer · 2017-10-06 06:23:22Z

16

Edit: Quoting from: How to return array of string with mongodb aggregation

The .aggregate() method always returns Objects no matter what you do and that cannot change.

Original Answer:
Try the aggregation framework:

db.myCol.aggregate([
    {
        $group:{_id:null, array:{$push:"$_id"}}
    },
    {
        $project:{array:true,_id:false}
    }
])

edited Oct 6, 2017 at 6:23

answered Oct 6, 2017 at 6:09

Dushyant Bangal

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3 Comments

Chi Dov Over a year ago

thanks, for you answer. this is what i used to try too. but i want to know is there a way to return a array of object like we do the db.collections.distinct("_id");

Dushyant Bangal Over a year ago

aggregate will always return an object

Avi E. Koenig Over a year ago

``` var resultSet = db.getCollection("mycollection").aggregate([ //optional match I use { $match: { "someField": "target value" } }, { $group: { _id: null, ids: { $push: "$_id" } } }, { $project: { ids: true, _id: false } } ]).toArray() var ids = resultSet[0].ids print(ids[0]) ```

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