聚合/过滤Mongo结果

时间:2017-02-05 19:50:45

标签: mongodb mongoose mongodb-query

我有一个充满投注市场的集合markets。我在集合中的数据如下所示:

{ "team" : "Milan", "back" : 1.0, "lay" : 1.0, "btotal" : 7.0, "ltotal" : 6.0 },
{ "team" : "Chelsea", "back" : 1.0, "lay" : 1.0, "btotal" : 1.0, "ltotal" : 1.0 },    
{ "team" : "Arsenal", "back" : 1.0, "lay" : 1.0, "btotal" : 1.0, "ltotal" : 1.0 },
{ "team" : "Milan", "back" : 3.0, "lay" : 3.0, "btotal" : 1.0, "ltotal" : 18.0 }

我想将我的数据显示为唯一名称列表,该行的最高btotal和相应的back赔率,最高ltotal及其对应的{{1}赔率。 从上面的例子中,返回的数据应为:

lay

这是第一次出现{"team":"Milan", "back":1.0, "lay":3.0, "btotal":7.0, "ltotal":18.0}, {"team":"Chelsea", "back":1.0, "lay":1.0, "btotal":1.0, "ltotal":1.0}, {"team":"Arsenal", "back":1.0, "lay":1.0, "btotal":1.0, "ltotal":1.0} btotal中的输出大于第二次出现的输出,因此返回"team":"Milan"赔率,{{1} } {第二次出现的back高于第一次,因此返回该行的ltotal赔率。

有关如何在Mongo或Mongoose中构建此查询的任何想法将不胜感激。

1 个答案:

答案 0 :(得分:0)

你能使用mapReduce吗?

地图功能

var map = function() {
    emit(this.team, this);
}

和reduce函数:

var reduce = function(key, values) {
    var res = values[0];
    for (var i = 1; i < values.length; i++) {
            if (values[i].btotal > res.btotal) {
                    res.back = values[i].back;
                    res.btotal = values[i].btotal;
            }
            if (values[i].ltotal > res.ltotal) {
                    res.lay = values[i].lay;
                    res.ltotal = values[i].ltotal;
            }
    }
    return res;
}

执行map-reduce操作:

db.markets.mapReduce(map, reduce, {out: "mpmarkets"});

使用$ project(聚合)生成预期输出:

cursor = db.mpmarkets.aggregate([{
    $project: {_id:0, 
    team: "$_id", 
    back: "$value.back",
    lay: "$value.lay",
    btotal: "$value.btotal",
    ltotal: "$value.ltotal"} },
    {$sort: {team: -1}
}]);

打印结果:

while ( cursor.hasNext() ) {
    printjson( cursor.next() );
}

结果:

{ "team" : "Milan", "back" : 1, "lay" : 3, "btotal" : 7, "ltotal" : 18 }
{ "team" : "Chelsea", "back" : 1, "lay" : 1, "btotal" : 1, "ltotal" : 1 }
{ "team" : "Arsenal", "back" : 1, "lay" : 1, "btotal" : 1, "ltotal" : 1 }
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