Question

我有一个具有以下结构的集合

{
    "_id" : "000aaa111",
    "reviewer_id" : "KQ5E8vKN5ezzgwH3C",
    "reviewee_id" : "FodX5fCg3RBe92Dpu",
    "stars" : [
        {
            "val" : "4.5",
            "for" : "communication"
        }, 
        {
            "val" : "4",
            "for" : "cleanliness"
        }
    ]
}

这实际上是为我的访问者给他们明星（1到5）的项目评分。那么我如何能够汇总我在集合中的所有文档来总结收到的所有明星。这是我期望的输出

{
    "reviewee_id": "FodX5fCg3RBe92Dpu",
    "overall": "5",
    "communication": "3",
    "cleanliness": "3.5"
}

Mongodb对我来说还是一个新手，我找不到任何与我现在的结构相匹配的例子。

你认为我的好去吗？

Answer 1

根据预期的输出，您需要调整结构以充分利用聚合。

你的结构应该是这样的。

<强>输入：

{
    "_id": ObjectId("585dc520772c68234445f2a4"),
    "reviewer_id": "KQ5E8vKN5ezzgwH3C",
    "reviewee_id": "FodX5fCg3RBe92Dpu",
    "stars": [{
        "communication": 4.5
    }, {
        "cleanliness": 4
    }, {
        "overall": 3
    }]
} {
    "_id": ObjectId("585dc520772c68234445f2a2"),
    "reviewer_id": "KQ5E8vKN5ezzgwH3D",
    "reviewee_id": "FodX5fCg3RBe92Dpu",
    "stars": [{
        "communication": 2.5
    }, {
        "cleanliness": 3
    }, {
        "overall": 4
    }]
}

<强>查询：

aggregate(
    [{
        $unwind: "$stars"
    }, {
        $group: {
            _id: "$reviewee_id",
            communication: {
                $avg: "$stars.communication"
            },
            cleanliness: {
                $avg: "$stars.cleanliness"
            },
            overall: {
                $avg: "$stars.overall"
            }
        }
    }]
)

<强>输出

{ "_id" : "FodX5fCg3RBe92Dpu", "communication" : 3.5, "cleanliness" : 3.5, "overall" : 3.5 }

在每个文档

1 个答案: