mongodb get distinct records

Question

I am using mongoDB in which I have collection of following format.

{"id" : 1 , name : x  ttm : 23 , val : 5 }
{"id" : 1 , name : x  ttm : 34 , val : 1 }
{"id" : 1 , name : x  ttm : 24 , val : 2 }
{"id" : 2 , name : x  ttm : 56 , val : 3 }
{"id" : 2 , name : x  ttm : 76 , val : 3 }
{"id" : 3 , name : x  ttm : 54 , val : 7 }

On that collection I have queried to get records in descending order like this:

db.foo.find({"id" : {"$in" : [1,2,3]}}).sort(ttm : -1).limit(3)

But it gives two records of same id = 1 and I want records such that it gives 1 record per id.

Is it possible in mongodb?

Answer 1

There is a distinct command in mongodb, that can be used in conjunction with a query. However, I believe this just returns a distinct list of values for a specific key you name (i.e. in your case, you'd only get the id values returned) so I'm not sure this will give you exactly what you want if you need the whole documents - you may require MapReduce instead.

Documentation on distinct: http://www.mongodb.org/display/DOCS/Aggregation#Aggregation-Distinct