How to find the document in mongodb where array size is greater than 1

Question

I want to find all the documents which are present and have array size greater than 1

My MongoDB collection looks like

{
  "_id" : ObjectId("5eaaeedd00101108e1123452"),
  "type" : ["admin","teacher","student"]
}
{
  "_id" : ObjectId("5eaaeedd00101108e1123453"),
  "type" : ["student"], 
}

How I find the document which has more than 1 type

Answer 1

You can do something like this. This is working version > 4.2

db.collection.find({
      $expr: {
        $gt: [
          {
            $size: "$type"
          },
          1
        ]
      }
    })

Working Mongo playground

If you use less, you can do something like follwoing

db.collection.find({
  type: {
    $gt: {
      $size: 1
    }
  }
})

Answer 2

The only working solution for this problem is as follows:

db.collection.find({
  $expr: {
    $gt: [
      {
        $size: "$arrayfield"
      },
      1
    ]
  }
})

All other solutions do not work. Tried it.

Answer 3

An alternative way to tackle the problem is to ask:

how can I find the document in MongoDB that has an element in the (n)-th position in the array field?

To achieve this, you can use the following query

{ "type.1": { $exists: true } }

this solution is cleaner and it does not require the use of aggregate pipelines

Answer 4

1. You can use $gt

db.collectionName.find({"type": {$gt: 1} });

2. You can use $where

db.collectionName.find( { $where: "type > 1" } );