MongoDB最大值查询和_id

Question

db.collections.find()

    { "_id" : ObjectId("55b0c2a0339bf8d00ab0bade"), "score" : 46, "playerid" : "45"}
    { "_id" : ObjectId("55b0c2de339bf8d00ab0badf"), "score" : 88, "playerid" : "45"}
    { "_id" : ObjectId("55b0cbca17f398f4281ab931"), "score" : 46, "playerid" : "99"}
    { "_id" : ObjectId("55b15ababe2df0f430d1cb93"), "score" : 89, "playerid" : "45"}

What I'm trying to do, is retrieve all documents with the largest score. If the same player occurs more than once, then we get the document with the largest score for that particular player.

The result would look like this:

    { "_id" : "55b0cbca17f398f4281ab931", "score" : 46 }
    { "_id" : "55b15ababe2df0f430d1cb93", "score" : 89 }

This is where I'm stuck:

db.players.aggregate([

{ "$group": { 
    "_id": "$playerid",
    score: { $max: "$score" }
} }
])

which returns:

    { "_id" : "99", "score" : "46" }
    { "_id" : "45", "score" : "89" }

Now, this is correct. But I just need the ObjectID instead.

Answer 1

Instead of $max then use $sort and $first where the other properties are important to you:

db.players.aggregate([
   { "$sort": { "score": -1 } },
   { "$group": { 
       "_id": "$playerid",
       "docId": { "$first": "$_id" },
       "score": { "$first": "$score" }
   }}
])

The $max operator of course only works on "one" field that you specify. In order to get detail from the collection document then you need to $sort and get the $first occurance on the grouping boundary.

Of course $first is relative to the $sort order which is "descending", otherwise use $last with ascending order, for the "maximum" value on the sort key.