Group Mongo documents by id and get the latest document by timestamp

Question

Imagine we have the following set of documents stored in mongodb:

{ \"fooId\" : \"1\", \"status\" : \"A\", \"timestamp\" : ISODate(\"2016-01-01T00:00:00.000Z         


        

Answer 1

You can use the $$ROOT system variable with the $last operator to return the last document.

db.collectionName.aggregate([      
    { "$sort": { "timestamp": 1 } },     
    { "$group": { 
        "_id": "$fooId",   
        "last_doc": { "$last": "$$ROOT" } 
    }}
])

Of course this will the last document for each group as a value of a field.

{
        "_id" : "2",
        "doc" : {
                "_id" : ObjectId("570e6df92f5bb4fcc8bb177e"),
                "fooId" : "2",
                "status" : "B",
                "timestamp" : ISODate("2016-01-02T00:00:00Z")
        }
}

---

If you are not happy with that output then your best bet will be to add another $group stage to the pipeline when you simply return an array of those documents using the $push accumulator operator.

db.collectionName.aggregate([      
    { "$sort": { "timestamp": 1 } },     
    { "$group": { 
        "_id": "$fooId",   
        "last_doc": { "$last": "$$ROOT" } 
    }},
    { "$group": { 
        "_id": null, 
        "result": { "$push": "$last_doc" } 
    }}

])