CODEWITHSUNDEEP
mongodbmongooseaggregation-framework
Steve Ng
Steve Ng

Reputation: 1189

Mongoose aggregate: How do I get all document that does not have a column populated?

This is my mongoose schema:

var userSchema = new mongoose.Schema({
  userID: String,
  gender: String,
  dateCreated: { type: Date, default: Date.now }
})

And this is my query

userSchema.statics.getUser = function(callback){
  this.aggregate()
    .exec(function(err, result){
      callback(result);
    });
}

Some users might not have gender added when creating, how can I pull all document using aggregate that does not have the attribute gender populated?

Upvotes: 0

Views: 1908

Answers (2)

Neil Lunn
Neil Lunn

Reputation: 151112

The results are actually language agnostic so a simple mongo shell should suffice. The results can simply be obtained in a find:

Consider:

> db.user.find().pretty()
{
    "_id" : ObjectId("52ecf6b46a8b205baeb2cc58"),
    "user" : "neil",
    "gender" : "M"
}
{ "_id" : ObjectId("52ecf6c66a8b205baeb2cc59"), "user" : "fabio" }
{
    "_id" : ObjectId("52ecf77c6a8b205baeb2cc5a"),
    "user" : "markus",
    "gender" : "M",
    "preference" : "any"
}

This works well for:

> db.user.find({ preference: null }).pretty()
{
    "_id" : ObjectId("52ecf6b46a8b205baeb2cc58"),
    "user" : "neil",
    "gender" : "M"
}
{ "_id" : ObjectId("52ecf6c66a8b205baeb2cc59"), "user" : "fabio" }

And also with $match in aggregate:

> db.user.aggregate([{$match: { prefernce: null }}])
{
    "result" : [
            {
                    "_id" : ObjectId("52ecf6b46a8b205baeb2cc58"),
                    "user" : "neil",
                    "gender" : "M"
            },
            {
                    "_id" : ObjectId("52ecf6c66a8b205baeb2cc59"),
                    "user" : "fabio"
            },
            {
                    "_id" : ObjectId("52ecf77c6a8b205baeb2cc5a"),
                    "user" : "markus",
                    "gender" : "M",
                    "preference" : "any"
            }
    ],
    "ok" : 1
}

Look at Aggregation in the MongoDB docs for more information.

Upvotes: 3

Peter Lyons
Peter Lyons

Reputation: 146014

You don't need aggregate for this. Just a regular old query will do.

User.find({gender: {$exists: false}}, function (error, users) {/*...*/}};

Upvotes: 0

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