分组在mongoDB中

Question

我将尝试尽我所能解释自己，我有这个系列，我想通过＆＃34;来执行一个＆＃34;组。我的收藏品的方式是......有一个看起来像这样的集合：

[
  {
    "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..}    
  }
]

输出就像这样

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

我已经做过像这样的小组

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

但是我无法找到一种方法来为每个结果添加密钥＆＃34; population＆＃34;。 （人口和城市的价值不会发生变化，我的意思是，如果城市的一个价值是＆＃34; city1＆＃34;，人口的价值总是很多＆＃34;太多＆＃34;）

Answer 1

使用mongo aggregation和$exists查找city礼物与否，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()

修改

根据您的第二个要求，请考虑您的文档结构：

{ "_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()

新修改

根据this问题要求，如果你的文件是这样的

{ "_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" } }

然后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()