如何通过此示例在 mongoosejs 中使用聚合

Question

我是来问一个关于mongo聚合函数实现这个例子的问题。

场景

我有 3 个 Mongo 模式，即房子、家庭和教育，它们是：

House: {
_id: ObjectId,
state: Number,
houseNumber: Number
}

Family: {
    _id: ObjectId,
    houseId: ObjectId,//ref: house
    name: String,
    gender: String
}

Education: {
    _id: ObjectId,
    familyId: ObjectId,//ref: Family
    level: String, //might be one of ["primary","secondary","higher_secondary"]
}

预期输出：

{
    state1: {
        primary: {
            male: 3,
            female: 4   
        },
        secondary: {
            male: 4,
            female: 8
        }
    },
    state2: {
        primary: {
            male: 5,
            female: 4   
        },
        secondary: {
            male: 4,
            female: 6
        }
    }
}

我想按性别和病房对所有教育水平进行分组。

我做了什么：

我是 mongo 世界的新手，最近从 sql 转到 no-sql。我已经这样做了：

let edu = await Education.find({level: "primary"}).populate({
      path: "family",
      match: {gender: "male"},
      select: "house",
      populate: {
        path: "house",
        match: {state: 1},
        select: "_id"
      }
    });
let count = (await edu.filter(each => !isEmpty(each.family) && !isEmpty(each.family.house)).length) || 0;

通过这样做，我得到了从状态 1 学习小学的男性成员的数量。但我不能为每个数据一个一个地调用这个函数。

根据要求，样本数据为：

house = [
  {
    _id: AA1,
    state: 1,
    houseNumber: 101
  },
  {
    _id: AA2,
    state: 1,
    houseNumber: 102
  },
  {
    _id: AA3,
    state: 2,
    houseNumber: 201
  }
];

family = [
  {
    _id: BB1,
    houseId: AA1, //ref: house
    name: "John",
    gender: "male"
  },
  {
    _id: BB2,
    houseId: AA1, //ref: house
    name: "Solena",
    gender: "female"
  },
  {
    _id: BB3,
    houseId: AA2, //ref: house
    name: "Efrain",
    gender: "male"
  },
  {
    _id: BB4,
    houseId: AA3, //ref: house
    name: "Naruto",
    gender: "male"
  }
];


education = [
  {
    _id: CC1,
    familyId: AA1, //ref: Family
    level: "primary"
  },
  {
    _id: CC2,
    familyId: AA2, //ref: Family
    level: "secondary"
  },
  {
    _id: CC3,
    familyId: AA3, //ref: Family
    level: "primary"
  },
  {
    _id: CC4,
    familyId: AA4, //ref: Family
    level: "secondary"
  }
];

P.S 预期输出与样本数据无关。并且 ObjectId 已被替换为一些唯一的引用。

这里有什么线索吗？

Answer 1

您可以在 4.x 版本中使用以下聚合查询。

查询家庭收藏并加入教育收藏获得等级，然后加入房屋收藏获得状态。

一旦您拥有按州、级别和性别划分的所有数据组，就可以计算所有匹配项，然后是其他组以格式化结果。最后阶段将聚合结果推广到自己的文档中。

最后两组将上一阶段的结果格式化为命名的键值文档。第一组将结果格式化为按州分组的性别和计数。第二组格式化先前组合的性别并使用教育键进行计数。

最后用状态键替换根阶段以格式化组合的性别、计数和教育文档。

为了清晰起见，还在每个阶段之后添加了输出。

db.family.aggregate(
[
  {"$lookup":{
    "from":"education",
    "localField":"_id",
    "foreignField":"familyId",
    "as":"education"
  }},
  {"$unwind":"$education"},
  {"$lookup":{
    "from":"house",
    "localField":"houseId",
    "foreignField":"_id",
    "as":"state"
  }},
  {"$unwind":"$state"},
  {"$group":{
    "_id":{
      "state":"$state.state",
      "education":"$education.level",
      "gender":"$gender"
    },
    "count":{"$sum":1}
  }},//{"_id":{"state" :1,"education" :"secondary","gender":"female"},"count":1}
  {"$group":{
    "_id":{"state":"$_id.state","education":"$_id.education"},
    "gandc":{"$mergeObjects":{"$arrayToObject":[[["$_id.gender","$count"]]]}}
  }},//{"_id":{"state":1,"education":"primary"},"gandc":{"male":2}}
  {"$group":{
    "_id":"$_id.state",
    "egandc":{"$mergeObjects":{"$arrayToObject":[[["$_id.education","$gandc"]]]}}
  }},//{"_id":1,"egandc":{"primary":{"male":2},"secondary":{"female":1}}}
  {"$replaceRoot":{"newRoot":{"$arrayToObject":[[[{"$toString":"$_id"},"$egandc"]]]}}} ])
])
//{"1":{"primary":{"male" : 2 },"secondary":{"female":1}}}