aggregation framework - MongoDB - Query - Top Document per Group -


this question has answer here:

i have following sql (mysql) query, , convert mongodb. getting top 1 document in every group:

select a.*  ads inner join (select id              ads             userid = 1             group ad_code             having max(ad_timestamp)) b on b.id = a.id

from have read far there several ways aggregate data in mongodb (more info @ mongodb aggregation comparison: group(), $group , mapreduce):

  • group (does not work on sharded collections)
  • mapreduce
  • $group

i trying solve mongodb aggregation framework. far have this:

db.ads.aggregate([     { $match: { userid: objectid("5976e215769d8a4a4d75c514") } },     {          $group: {              _id: "$ad_code",              latesttimestamp: { $max: "$ad_timestamp" },         }     } ])

but not return _ids of matching documents, ad code , max timestamp, cannot use data whole documents.

this question looks relevant, doesn't seem solve same issue i'm having: query 1 document per association mongodb

thanks

edit duplicate flag: question different others, , different question linked, in solution uses $first operand find 1-match, , docids field retrieve un-aggregated original documents. different how other questions have approached solutions, partly result of evolution of mongodb on time.

it doesn't returns _id because didn't asked it. try query instead : 

db.ads.aggregate([     { $match: { userid: objectid("5976e215769d8a4a4d75c514") } },     { $sort: {"ad_timestamp": -1}},     { $group: {          _id: "$ad_code",          latesttimestamp: { $first: "$ad_timestamp" },         docids: {$first: "$_id"}        }     } ])

here first sort ad_timestamp , first of each group $first, , same _id


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