Mongodb find a document with all subdocuments satisfying a condition

Question

I have Game collection in my DB:

var game = {
  players: [{username:\"user1\", status:\"played\"},
            {username:\"user2\", status:\"accepted\"}]
}
<         


        

Answer 1

If you only have one other status than "played" use the query:

db.games.find({ "players.status": { $ne:"accepted" } })

You can adjust the query to handle more status values, as long as they are all known at the time of the query.

Answer 2

Use the $unwind feature of the aggregation framework:

db.collection.aggregate([{$unwind:"$players.status"},{$match:{"players.status":"played"},{$project:{"_id":0,"players.status":1}}])

(check out this answer: https://stackoverflow.com/a/15082321/1214167)

Answer 3

I ran into this problem too. I was trying to use $all and making little to no progress. It helped to flip my perspective on the problem and use $not instead. I wrote about what I found here.

In your situation, you should be able to do something like this:

db.games.find({
  players: {
    $not: {
      $elemMatch: {
        status: { $ne: "played" }
      }
    }
  }
})

Find all documents with a player status not equal to "played" and return the inverse of that set. The nice thing about this is that you never have to worry about adding more status values.