javascript - mongoDB 中的分组依据

Question

我会尽力解释自己，我有这个集合，我想以某种方式对我的集合执行“分组依据”......拥有一个如下所示的集合:

[
  {
    "city": "City1",
    "population": "too many",
    "person": {//..some object..}   
  },
  {
    "city": "City1",
    "population": "too many",
    "person": {//..other object..}  
  },
  {
    "city": "City1",
    "population": "too many",
    "person": {//..another object..}    
  },
  {
    "city": "City2",
    "population": "too low",
    "person": {//..one object..}    
  }
]

输出会是这样

[
  {
    "city": "City1",
    "population": "too many",
    "person": [
       {//..some object..},
       {//..other object..},
       {//..another object..}          
     ]
  },
  {
    "city": "City2",
    "population": "too low",
    "person":[ {//..one object..}   ]
  }
]

我已经准备好了像这样的groupBy

db.myCollection.aggregate({
     "$group": {
     "_id": "$city",
        "resources": {
        "$push": "$person"
     }
  }
})

但我找不到将关键“population”添加到每个结果的方法。(人口和城市的值不会改变，我的意思是如果城市的一个值为“city1”，则人口的值将始终“太多”)

Answer 1

使用mongo aggregation和 $exists查找 city 是否存在并查询如下:

 db.collectionName.aggregate({
   "$match": {
     "city": {
       "$exists": true //check city presents or not
     }
   }
 }, {
   "$group": {
     "_id": "$city", // group by city
     "data": {
       "$push": { // push all data into array 
         "name": "$name",
         "age": "$age",
         "address": "$address"
       }
     }
   }
 }, {
   "$project": {
     "city": "$_id",
     "theResults": "$data", //project them 
     "_id": 0
   }
 }).pretty()

编辑

根据您的第二个要求，考虑您的文档结构如下:

{ "_id" : ObjectId("5564aaa1934833d8c1a1313f"), "city" : "City1", "person" : { "name" : "someName", "age" : "32", "address" : "Evergreen 124" } }
{ "_id" : ObjectId("5564aaa1934833d8c1a13140"), "city" : "City1", "person" : { "name" : "someName", "age" : "32", "address" : "Evergreen 125" } }
{ "_id" : ObjectId("5564aaa1934833d8c1a13141"), "city" : "City1", "person" : { "name" : "someName", "age" : "32", "address" : "Evergreen 126" } }
{ "_id" : ObjectId("5564aaa1934833d8c1a13142"), "city" : "City2", "person" : { "name" : "someName", "age" : "32", "address" : "Evergreen 129" } }

你应该使用这个聚合:

db.collectionName.aggregate({
  "$group": {
    "_id": "$city", //group by city
    "data": {
      "$push": {
        "name": "$person.name",
        "age": "$person.age",
        "address": "$person.address"
      }
    }
  }
}, {
  "$project": {
    "_id": 0,
    "city": "$_id",
    "person": "$data"
  }
}).pretty()

新编辑

根据 this如果您的文件是这样的，请询问要求

{ "_id" : ObjectId("5564aaa1934833d8c1a1313f"), "city" : "City1", "population" : "too many", "person" : { "name" : "someName", "age" : "32", "address" : "Evergreen 124" } }
{ "_id" : ObjectId("5564aaa1934833d8c1a13140"), "city" : "City1", "population" : "too many", "person" : { "name" : "someName", "age" : "32", "address" : "Evergreen 125" } }
{ "_id" : ObjectId("5564aaa1934833d8c1a13141"), "city" : "City1", "population" : "too many", "person" : { "name" : "someName", "age" : "32", "address" : "Evergreen 126" } }
{ "_id" : ObjectId("5564aaa1934833d8c1a13142"), "city" : "City2", "population" : "too low", "person" : { "name" : "someName", "age" : "32", "address" : "Evergreen 129" } }

那么你应该使用以下聚合:

db.collectionName.aggregate({
  "$group": {
    "_id": {
      "city": "$city",
      "population": "$population"
    },
    "data": {
      "$push": {
        "name": "$person.name",
        "age": "$person.age",
        "address": "$person.address"
      }
    }
  }
}, {
  "$project": {
    "_id": 0,
    "city": "$_id.city",
    "population": "$_id.population",
    "person": "$data"
  }
}).pretty()