Query for documents where array size is greater than 1

Question

I have a MongoDB collection with documents in the following format:

{
  \"_id\" : ObjectId(\"4e8ae86d08101908e1000001\"),
  \"name\" : [\"Name\"],
  \"zipcod         


        

Answer 1

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" ] }
>

Answer 2

Update:

For mongodb versions 2.2+ more efficient way to do this described by @JohnnyHK in another answer.

---

1.Using $where

db.accommodations.find( { $where: "this.name.length > 1" } );

But...

Javascript executes more slowly than the native operators listed on this page, but is very flexible. See the server-side processing page for more information.

2.Create extra field NamesArrayLength, update it with names array length and then use in queries:

db.accommodations.find({"NamesArrayLength": {$gt: 1} });

It will be better solution, and will work much faster (you can create index on it).

Answer 3

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}}})

Answer 4

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]}})

Answer 5

There's a more efficient way to do this in MongoDB 2.2+ now that you can use numeric array indexes 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}}}
);

Answer 6

I know its old question, but I am trying this with $gte and $size in find. I think to find() is faster.

db.getCollection('collectionName').find({ name : { $gte : {  $size : 1 } }})