Javascript mongoDB中的分组_Javascript_Mongodb_Mongoose

Javascript mongoDB中的分组

javascript mongodb mongoose

Javascript mongoDB中的分组,javascript,mongodb,mongoose,Javascript,Mongodb,Mongoose,我会尽我所能解释我自己，我有这个收藏，我想对我的收藏进行一次“分组”。。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..} }, {

我会尽我所能解释我自己，我有这个收藏，我想对我的收藏进行一次“分组”。。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..}    
  }
]

输出是这样的吗

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

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

我已经准备好做这样的群比了

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

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

但我找不到一种方法将关键的“总体”添加到每个结果中。（人口和城市的值不变，我的意思是如果城市的一个值是“city1”，人口的值总是“太多”）

使用mongo，查找城市是否存在，并按如下方式查询：

 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

根据你的第二个要求，考虑你的文档结构如下：

{ "_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" } }

您应该使用此聚合：

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()

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()

新编辑

根据问题要求，如果您的文件像这样

{ "_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，您应该使用以下聚合：

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()

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()

使用mongo和查找是否存在城市，并按如下方式进行查询：

 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

根据你的第二个要求，考虑你的文档结构如下：

{ "_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" } }

您应该使用此聚合：

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()

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()

新编辑

根据问题要求，如果您的文件像这样

{ "_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，您应该使用以下聚合：

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()

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()

这与我想做的很接近，但是如果集合像这样：

code

[{“城市”：“城市1”，“人”：{/..some object….}，{“城市”：“城市1”，“人”：{/..other object….}，{“城市”：“城市1”，“人”：{/..other object….}，{“城市2”{/，“person”：{/..one object..}}]

code

@flaalf我不明白你在问什么？你的收藏有不同的数据吗？这接近于我想做的，但是如果像这样的收藏：

code

[{“城市”：“City1”，“person”：{/..some object..}，{“城市”，会有多大的不同“：”City1“，”person“：{/..other object..}，{”City1“，”person“，{/..other object..}，{”city“：”City2“，”person“：{/..one object..}}]

code

@flaalf我不明白你在问什么？您的收藏是否有不同的数据？