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我有以下MongoDB集合“ Games”：

{ 
"_id" : ObjectId("515461d3c6c18efd4a811fd3"), 
"gameid" : NumberLong("86982207656"), 
"tableName" : "Hydra Zoom 40-100 bb", 
"nplayers" : 6, 
"playersList" : [   
                                    {           "exist" : true,         
                                                        "suspended" : false,
                                                        "grade" : 0,
                                                        "clusterId" : -1,
                                                        "playerid" : "DoomY9999",
                                                        "playsWithFriends" : 0,
                                                        "squeezePlay" : 0,
                                                        "weakShowdown" : 0,
                                                        "numberOfPlays" : 1
                                     },
                                  {
                                                        "exist": true,
                                                        "suspended" : false,

我想将以下MySQL查询映射到 MongoDB

String query = "SELECT idplayer, COUNT(idplayer) AS countplayer "
                    + "FROM (SELECT b.idgame, b.idplayer "
                    + "FROM associations a, associations b "
                    + "WHERE a.idplayer=? "
                    + "AND b.idgame=a.idgame "
                    + "AND b.idplayer <> ? "
                    + "ORDER BY b.idgame DESC LIMIT 1000) as c"
                    + " GROUP BY idplayer "
                    + "ORDER BY countplayer DESC LIMIT 5;";

查询描述：此 SQL 查询计算出现与玩家“X”玩同一游戏的最频繁玩家。结果将是玩家的名字和一起玩的次数

LIMIT 的简短说明：第一个“ LIMIT 1000”实际上是我们要检查的游戏的限制，因为数据库可能非常大，我们只分析按 DESC 顺序排列的最后 1000 场游戏（最近有更高的“ gameid”）。

第二个限制5：用于“前5”的朋友。我们将总结他们的人数。

到目前为止，我已经完成了：几乎所有使用聚合框架的东西，都为"ORDER BY b.idgame DESC LIMIT 1000) as c". 这对我来说很重要，因为它经历的游戏数量可能非常高。

这是我在MongoDB（Java 驱动程序）中的查询： 

 //build the query
    DBObject match1 = new BasicDBObject("$match", new BasicDBObject("playersList.playerid",_playerid));
    DBObject unwind = new BasicDBObject("$unwind", "$playersList");
    DBObject match2 = new BasicDBObject("$match", new BasicDBObject("playersList.playerid",new BasicDBObject("$ne",_playerid)));

    DBObject groupFields = new BasicDBObject("_id","$playersList.playerid");
    groupFields.put("times", new BasicDBObject("$sum",1));
    DBObject group = new BasicDBObject("$group", groupFields);
    DBObject sort = new BasicDBObject("$sort", new BasicDBObject("times",-1) );
    DBObject limit = new BasicDBObject("$limit", 5 );

    DBObject group2 = new BasicDBObject("$group", "gameid");
    DBObject sort2 = new BasicDBObject("$sort", new BasicDBObject("gameid",-1) );
    DBObject limit2 = new BasicDBObject("$limit", 1000 );

    DB db = mongoDb;
    DBCollection coll = db.getCollection("games");

    //aggregation query
    //THIS WORKS
    AggregationOutput output = coll.aggregate( match1, unwind, match2, group, sort, limit);

    //THIS DOESN'T WORK!
     AggregationOutput output = coll.aggregate( match1, unwind, match2, group, sort, limit, group2, sort2, limit2);

请帮我解决这个查询。谢谢！

4

1 回答 1

2

第一次操作后该字段game不在结果中，因此基于该字段的第二次操作将不起作用。groupgroupgame

为了更有效的查询，您应该重新排序聚合操作以尽早减少数据。我在展开 playerList 之前移动了比赛的匹配，没有必要拥有第二组。

聚合操作在 mongo shell 中是这样的：

// playerId to search for coplayers
var playerId = "DoomY9999"

db.game.aggregate([
    // First $match can take advantage of suitable index if available

    // Find all games that playerid X has played
    { $match : { "playersList.playerid" : playerId } },

    // Sort by most recent games (gameid descending)
    { $sort : { "_id.gameid" : -1 } },

    // Limit number of games to examine
    { $limit : 1000 },

    // Create a stream of documents from the playersList array
    { $unwind : "$playersList" },

    // Match players except for playerid X
    { $match : { "playersList.playerid" : {$ne : playerId }} },

    // Count number of games each player has played
    { $group : {
        _id : "$playersList.playerid",
        count : { $sum : 1 }
    }},

    // Sort by most frequent players (count descending)
    { $sort : { "count" : -1 } },

    // Limit results to 5 players
    { $limit : 5 },

    // Rename the result fields
    { $project : {
        _id : 0,
        coplayer : "$_id",
        count : 1
    }}
])
于 2013-04-12T04:55:36.560 回答