mongodb 中两个字段的总出现次数

sum number of occurrences on two fields in mongodb

我有一个叫games的文档合集,长这样

{ winner: 'oakis', loser: 'test' },
{ winner: 'test', loser: 'oakis' }

现在我想MongoDB输出这个:

{ _id: 'oakis', wins: 1, losses: 1 },
{ _id: 'test', wins: 1, losses: 1 }

我一直在研究聚合、mapReduce 和 distinct,但没有成功。 我希望你能帮助我!

编辑:

我有一个名为 "tied" 的字段,它也是一个布尔值。我怎样才能在结果中实现绑定游戏? (游戏中的每个玩家也有一个字段,player1 & player2) 结果应该是: { _id: 'name', wins: 3, losses: 1, tied: 1 }

编辑 2:

这样解决了!

db.games.aggregate([
    { 
        $project: { 
            scores: [
                { name: '$winner', wins: { $literal: 1 }, losses: { $literal: 0 }, tied: { $literal: 0 } }, 
                { name: '$loser', wins: { $literal: 0 }, losses: { $literal: 1 }, tied: { $literal: 0 } },
        { name: '$player1', wins: { $literal: 0 }, losses: { $literal: 0 }, tied: { $cond: [ "$tied", 1, 0 ] } },
        { name: '$player2', wins: { $literal: 0 }, losses: { $literal: 0 }, tied: { $cond: [ "$tied", 1, 0 ] } }
            ] 
        } 
    }, 
    { 
        $unwind: '$scores' 
    }, 
    { 
        $group: {
             _id: "$scores.name", 
            wins: { $sum: "$scores.wins" }, 
            losses: { $sum: "$scores.losses" },
    tied: { $sum: "$scores.tied" }
        } 
    }
])

借助聚合框架,您可以使用如下所示的管道。本质上,它首先将数据投影到 [{name: $winner, wins: 1, losses: 0}, {name: $loser, wins: 0, losses: 1}] 形式的 2 元素数组中,然后展开数组以生成单个元素数组,然后按名称分组,并总结出得失。

games.aggregate(
    [
        { 
            $project: { 
                scores: [
                    { name: '$winner', wins: { $literal: 1 }, losses: { $literal: 0 } }, 
                    { name: '$loser', wins: { $literal: 0 }, losses: { $literal: 1 } }
                ] 
            } 
        }, 
        { 
            $unwind: '$scores' 
        }, 
        { 
            $group: {
                 _id: "$scores.name", 
                wins: { $sum: "$scores.wins" }, 
                losses: { $sum: "$scores.losses" } 
            } 
        }
    ]
)