Query grouped by two swap fields - mongodb

Question

I have collection messages with the following documents
{
"_id" : ObjectId("5164218f359f109fd4000012"),
"receiver_id" : ObjectId("5164211e359f109fd4000004"),
"sender_id" : ObjectId("5162de8a359f10cbf700000c"),
"body" : "Hello Billy!!!",
"readed" : false,
"updated_at" : ISODate("2013-04-09T14:11:27.17Z"),
"created_at" : ISODate("2013-04-09T14:11:27.17Z")
}
I need to make query for receive last messages(don't matter recieved or sended) for a given user (grouped by reciever_id+sender_id fields) and sorted by created_at.
To better explain the question, an example of how I did it in SQL:
SELECT DISTINCT ON (sender_id+receiver_id) * FROM messages
ORDER by (sender_id+receiver_id), created_at DESC
WHERE sender_id = given_user or receiver_id = given_user
I don't understand how to solve this problem with mondodb.

Answer 1

The Aggregation Framework in MongoDB 2.2+ provides the most obvious translation of your query. The MongoDB manual includes an SQL to Aggregation Framework Mapping Chart as a general guide, although there are definite differences in the two approaches.
Here's a commented example you can try in the mongo shell:
var given_user = ObjectId("5162de8a359f10cbf700000c");
db.messages.aggregate(
// match: WHERE sender_id = given_user or receiver_id = given_user
// NB: do the match first, because it can take advantage of an available index
{ $match: {
$or:[
{ sender_id: given_user },
{ receiver_id: given_user },
]
}},
{ $group: {
// DISTINCT ON (sender_id+receiver_id)
_id: { sender_id: "$sender_id", receiver_id: "$receiver_id" }
}},
// ORDER by (sender_id+receiver_id), created_at DESC
{ $sort: {
sender_id: 1,
receiver_id: 1,
created_at: -1
}}
)
Sample result:
{
"result" : [
{
"_id" : {
"sender_id" : ObjectId("5162de8a359f10cbf700000c"),
"receiver_id" : ObjectId("5164211e359f109fd4000004")
}
}
],
"ok" : 1
}
You may want to add additional fields on the grouping, such as a count of messages received.
If you actually want to combine the sender_id+receiver_id into a single field, you can use the $concat operator in MongoDB 2.4+.

Answer 2

There is no explicit way to do so. Let's review workarounds:
Way 1:
do the distinct at code level (after find), then just use find:
db.message.find({$or:[{sender_id:?}, {receiver_id:?}]})
Way 2:Using aggregation framework :
db.message.aggregate( [
{$match: {$or:[{sender_id:?}, {receiver_id:?}]},
$group: { _id: {sender:"$sender_id", receiver:"$receiver_id"},
other: { ... } } },
$sort: {sender_id,receiver_id,...}
] )
This way problem appears at sort level since sender_id, receiver_id is not the same as sender_id+receiver_id
Way 3: Introduce the surrogate field sender_id+receiver_id then use find or even distinct per Stennie hint.