MongoDB query only the inner document - mongodb

Question

My mongodb collection looks like this:
{
"_id" : ObjectId("5333bf6b2988dc2230c9c924"),
"name" : "Mongo2",
"notes" : [
{
"title" : "mongodb1",
"content" : "mongo content1"
},
{
"title" : "replicaset1",
"content" : "replca content1"
}
]
}
{
"_id" : ObjectId("5333fd402988dc2230c9c925"),
"name" : "Mongo2",
"notes" : [
{
"title" : "mongodb2",
"content" : "mongo content2"
},
{
"title" : "replicaset1",
"content" : "replca content1"
},
{
"title" : "mongodb2",
"content" : "mongo content3"
}
]
}
I want to query only notes that have the title "mongodb2" but do not want the complete document.
I am using the following query:
> db.test.find({ 'notes.title': 'mongodb2' }, {'notes.$': 1}).pretty()
{
"_id" : ObjectId("5333fd402988dc2230c9c925"),
"notes" : [
{
"title" : "mongodb2",
"content" : "mongo bakwas2"
}
]
}
I was expecting it to return both notes that have title "mongodb2".
Does mongo return only the first document when we query for a document within a document ?

Answer 1

The positional $ operator can only return the first match index that it finds.
Using aggregate:
db.test.aggregate([
// Match only the valid documents to narrow down
{ "$match": { "notes.title": "mongodb2" } },
// Unwind the array
{ "$unwind": "$notes" },
// Filter just the array
{ "$match": { "notes.title": "mongodb2" } },
// Reform via group
{ "$group": {
"_id": "$_id",
"name": { "$first": "$name" },
"notes": { "$push": "$notes" }
}}
])
So you can use this to "filter" specific documents from the array.

Answer 2

$ always refers to the first match, as does the $elemMatch projection operator.
I think you have three options:
separate the notes so each is a document of its own
accept sending more data over the network and filter client-side
use the aggregation pipeline ($match and $project)
I'd probably choose option 1, but you probably have a reason for your data model.