Question

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

{
  "_id" : ObjectId("4e8ae86d08101908e1000001"),
  "name" : ["Name"],
  "zipcode" : ["2223"]
}
{
  "_id" : ObjectId("4e8ae86d08101908e1000002"),
  "name" : ["Another ", "Name"],
  "zipcode" : ["2224"]
}

I can currently get documents that match a specific array size:

db.accommodations.find({ name : { $size : 2 }})

This correctly returns the documents with 2 elements in the name array. However, I can’t do a $gt command to return all documents where the name field has an array size of greater than 2:

db.accommodations.find({ name : { $size: { $gt : 1 } }})

How can I select all documents with a name array of a size greater than one (preferably without having to modify the current data structure)?

Answer 1

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).