Question

让我们说我有3个模式

让我们说一个模式是

房屋模式
家庭模式
教育模式

// 1 house 1 schema
const HouseSchema = Schema({
    state: Number,
    houseNumber: String, 
    houseImage: String,
    houseAge: Number, 
});

const FamilySchema = Schema({
    house: {
        type: Schema.Types.ObjectId,
        ref: "house"
    },
    fullName: String,
    gender: String
});

const EducationSchema = new Schema({
    family: {
        type: Schema.Types.ObjectId,
        ref: "familydetail"
    },
    educationLevel: String,
    educationFinishedYear: Number
});

我想要这样一种结果，即所有按性别划分的家庭成员都生活在州：1 (ref: house)，结果组按educationLevel居住。

我通过使用填充和匹配功能（例如：

）简单地完成了此操作

const getEducationCount = async (education, sex, ward) => {
try {
    let edu = await Education.find(education, "family").populate({
        path: "family",
        match: sex,
        select: "house",
        populate: {
            path: "house",
            match: ward,
            select: "_id"
        }
    });

    if (edu) {
        // console.log("education", edu);
        let count =
            (await edu.filter(
                each => !isEmpty(each.family) && !isEmpty(each.family.house)
            ).length) || 0;
        // console.log(count);
        return count;
    }
} catch (error) {
    console.log(error);
}
};

我必须为每个educationlevel调用此函数，这需要很长时间。它始终扫描n number of documents。如何优化呢？

我该怎么做：

    Education.aggregate([
    {
        $match: {}
    },
    {
        $group: {
            _id: "$educationLevel",
        },  
    },
    {
        $lookup: {
            from: "family",
            localField: "family",
            foreignField: "_id",
            as: "family",
            $group: {
                _id: "$gender",
            },
            $lookup: {
                from: "house",
                localField: "house",
                foreignField: "_id",
                as: "house",
                $match: {
                    state: 1,
                }
            }
        }
    }
])

任何参考和提示将不胜感激。

谢谢。

MongoDB：如何在聚合函数中填充匹配项和分组依据

0 个答案: