Query for Documents Where Array Size Is Greater Than 1

There's a more efficient way to do this in MongoDB 2.2+ now that you can use numeric array indexes (0 based) in query object keys.

// Find all docs that have at least two name array elements.
db.accommodations.find({'name.1': {$exists: true}})

You can support this query with an index that uses a partial filter expression (requires 3.2+):

// index for at least two name array elements
db.accommodations.createIndex(
    {'name.1': 1},
    {partialFilterExpression: {'name.1': {$exists: true}}}
);

I believe this is the fastest query that answers your question, because it doesn't use an interpreted $where clause:

{$nor: [
    {name: {$exists: false}},
    {name: {$size: 0}},
    {name: {$size: 1}}
]}

It means "all documents except those without a name (either non existant or empty array) or with just one name."

Test:

> db.test.save({})
> db.test.save({name: []})
> db.test.save({name: ['George']})
> db.test.save({name: ['George', 'Raymond']})
> db.test.save({name: ['George', 'Raymond', 'Richard']})
> db.test.save({name: ['George', 'Raymond', 'Richard', 'Martin']})
> db.test.find({$nor: [{name: {$exists: false}}, {name: {$size: 0}}, {name: {$size: 1}}]})
{ "_id" : ObjectId("511907e3fb13145a3d2e225b"), "name" : [ "George", "Raymond" ] }
{ "_id" : ObjectId("511907e3fb13145a3d2e225c"), "name" : [ "George", "Raymond", "Richard" ] }
{ "_id" : ObjectId("511907e3fb13145a3d2e225d"), "name" : [ "George", "Raymond", "Richard", "Martin" ] }
>

You can use aggregate, too:

db.accommodations.aggregate(
[
     {$project: {_id:1, name:1, zipcode:1, 
                 size_of_name: {$size: "$name"}
                }
     },
     {$match: {"size_of_name": {$gt: 1}}}
])

// you add "size_of_name" to transit document and use it to filter the size of the name

You can use $expr ( 3.6 mongo version operator ) to use aggregation functions in regular query.

Compare query operators vs aggregation comparison operators.

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

Try to do something like this:

db.getCollection('collectionName').find({'ArrayName.1': {$exists: true}})

1 is number, if you want to fetch record greater than 50 then do ArrayName.50 Thanks.

MongoDB 3.6 include $expr

You can use $expr in order to evaluate an expression inside a $match, or find.

{ $match: {
           $expr: {$gt: [{$size: "$yourArrayField"}, 0]}
         }
}

or find

collection.find({$expr: {$gte: [{$size: "$yourArrayField"}, 0]}});

None of the above worked for me. This one did so I'm sharing it:

db.collection.find( {arrayName : {$exists:true}, $where:'this.arrayName.length>1'} )

db.accommodations.find({"name":{"$exists":true, "$ne":[], "$not":{"$size":1}}})

I found this solution, to find items with an array field greater than certain length

db.allusers.aggregate([
  {$match:{username:{$exists:true}}},
  {$project: { count: { $size:"$locations.lat" }}},
  {$match:{count:{$gt:20}}}
])

The first $match aggregate uses an argument thats true for all the documents. If blank, i would get

"errmsg" : "exception: The argument to $size must be an Array, but was of type: EOO"

Although the above answers all work, What you originally tried to do was the correct way, however you just have the syntax backwards (switch "$size" and "$gt")..

Correct:

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

Incorrect:

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

You can MongoDB aggregation to do the task:

db.collection.aggregate([
  {
    $addFields: {
      arrayLength: {$size: '$array'}
    },
  },
  {
    $match: {
      arrayLength: {$gt: 1}
    },
  },
])