How to find random record in Mongoose

Question

How can I find random records in MongoDB?

I found multiple articles here on StackOverflow, but I couldn't understand them. Like for example:

db.yourCollection.find().limit(-1).skip(yourRandomNumber).next()

How would I execute it in my code? (collection is User)

User.findOne(RANDOM PLAYER).then(result) {
    console.log(result);
}

im struggeling with Users.findOne().skip(rand).then(function(result) { @Yashua — maria, Sep 02 '16 at 07:48
possible duplicate of https://stackoverflow.com/questions/14644545/random-document-from-a-collection-in-mongoose — Deepak, Jul 31 '19 at 10:34

cyberwombat · Accepted Answer · 2020-12-21T05:01:07.927

75

The idea behind getting a random record is to query all the matching records but just get one. This is what findOne() does without any criteria given.

Then you will want to pick a random entry in all the possible matches. This is done by:

Find out how many possible entries there could be - we use count() on the collection for this. Note that, as mentioned in comments, count is deprecated in version 4 and one should use estimatedDocumentCount or countDocuments instead. The different lies in precision/memory usage amongst other things. Here's a SO post discussing it a bit.
Come up with a random number within our count.
Use skip() to "skip" to the desired match and return that.

Here's a snippet as modified from this SO answer:

// Get the count of all users
User.count().exec(function (err, count) {

  // Get a random entry
  var random = Math.floor(Math.random() * count)

  // Again query all users but only fetch one offset by our random #
  User.findOne().skip(random).exec(
    function (err, result) {
      // Tada! random user
      console.log(result) 
    })
})

edited Dec 21 '20 at 05:01

answered Sep 02 '16 at 16:44

cyberwombat

38,105
35
175
251

2

Exactlly what I was looking for. Thank you ! – Pramesh Bajracharya Aug 27 '17 at 08:42
nice work!! This is what I was looking for. – Eray T Nov 08 '18 at 16:26
4

count() - collection.count is deprecated, and will be removed in a future version. Use Collection.countDocuments or Collection.estimatedDocumentCount instead. – Prathamesh More Sep 30 '19 at 08:21
3

What about using `const testimonial = await Model.aggregate().sample(2)` ? [docs](https://mongoosejs.com/docs/api.html#aggregate_Aggregate-sample) – aasutossh Jul 08 '22 at 13:34

score 1 · Answer 2 · answered Oct 05 '16 at 12:22

1

To get random document(s) from mongodb by using mongoose.

    limitrecords=10;

    function getRandomArbitrary(min, max) {
      return Math.ceil(Math.random() * (max - min) + min);
    }

    var userschema = new Schema({
      name: String
    });

    User = mongoose.model('User', userschema);

    User.count({your_query},function(err,count){

       var skipRecords = getRandomArbitrary(1, count-limitrecords);

       query.skip(skipRecords); // Random Offset

       query.exec(function(err,result){
         console.log(result);  // 10 random users 
       });

    });

This is an example for 10 random records, you can set "limitrecords" according to your requirement.

Thanks!

answered Oct 05 '16 at 12:22

Rahul Thakkar

43
5

2

`ReferenceError: query is not defined` - where are you setting query from? – Luke Brown Jun 11 '19 at 11:20
Luke: https://mongoosejs.com/docs/api/query.html#query_Query That is a query. So something like `User.find()` or `User.where({ name: 'vonderful' })`. – Nebulosar Jul 13 '21 at 12:02
Heads up. That this answer returns a random start of the find but the sequence will always be the same. Original: [ 1 2 3 4 5 ] with limitrecords=3, possible returns: [ 1, 2, 3 ], [2 3 4 ],[ 3 4 5 ] – Ezequiel Santiago Sánchez Sep 06 '21 at 05:46

How to find random record in Mongoose

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