javascript - 使用聚合 $lookup 和 $mergeObjects

Question

我要加入收藏。以前，我只使用查找，这样我就可以得到连接的分离字段。但我需要得到类似 mysql join 的结果。我注意到此操作有 $lookup 和 $mergeObjects，但效果不佳。

用户集合模型。

{
    "_id": ObjectId("xxxxxxx"), //this is default id from mongoDB
    "name": 'admin user',
    "email": 'admin@test.com',
    "password": 'xxxxxxxx',
    "roles": [
        {
            "id": 0,
            "approved": true
        },{
            "id": 2,
            "approved": true
        }
    ]
},{
    "_id": ObjectId("xxxxxxx"), //this is default id from mongoDB
    "name": 'userOne',
    "email": 'user@test.com',
    "password": 'xxxxxxxx',
    "roles": [
        {
            "id": 1,
            "approved": true
        }
    ]
}

Angular 色集合模型。

{
    "_id": ObjectId("xxxxxxx"), //this is default id from mongoDB
    "id": '0',
    "name": 'administrator'
},{
    "_id": ObjectId("xxxxxxx"), //this is default id from mongoDB
    "id": '0',
    "name": 'employeer'
},{
    "_id": ObjectId("xxxxxxx"), //this is default id from mongoDB
    "id": '0',
    "name": 'freelancer'
} 

加入后，我想得到如下结果。

{
    "_id": ObjectId("xxxxxxx"), //this is default id from mongoDB
    "name": 'admin user',
    "email": 'admin@test.com',
    "password": 'xxxxxxxx',
    "roles": [
        {
            "id": 0,
            "name": "administrator",    //join result
            "approved": true
        },{
            "id": 2,
            "name": "freelancer",       //join result
            "approved": true
        }
    ]
},{
    "_id": ObjectId("xxxxxxx"), //this is default id from mongoDB
    "name": 'userOne',
    "email": 'user@test.com',
    "password": 'xxxxxxxx',
    "roles": [
        {
            "id": 1,
            "name": "employeer",        //join result
            "approved": true
        }
    ]
}

Answer 1

您可以使用下面的aggregation使用 mongodb 3.4

您需要$unwind首先是 Angular 色数组，然后是 $group再次回滚

db.users.aggregate([
  { "$unwind": "$roles" },
  { "$lookup": {
    "from": "roles",
    "localField": "roles.id",
    "foreignField": "id",
    "as": "roles.role"
  }},
  { "$unwind": "$roles.role" },
  { "$addFields": {
    "roles": { "$mergeObjects": ["$roles.role", "$roles"] }
  }},
  { "$group": {
    "_id": "$_id",
    "email": { "$first": "$email" },
    "password": { "$first": "$password" },
    "roles": { "$push": "$roles" }
  }},
  { "$project": { "roles.role": 0 }}
])

使用 mongodb 3.6 及以上版本非常简单

db.users.aggregate([
  { "$unwind": "$roles" },
  { "$lookup": {
    "from": "roles",
    "let": { "roleId": "$roles.id", "approved": "$roles.approved" },
    "pipeline": [
      { "$match": { "$expr": { "$eq": ["$id", "$$roleId"] }}},
      { "$addFields": { "approved": "$$approved" }}
    ],
    "as": "roles"
  }},
  { "$unwind": "$roles" },
  { "$group": {
    "_id": "$_id",
    "email": { "$first": "$email" },
    "password": { "$first": "$password" },
    "roles": { "$push": "$roles" }
  }}
])

两者都会给你相似的Output

[
  {
    "_id": ObjectId("5a934e000102030405000004"),
    "email": "user@test.com",
    "password": "xxxxxxxx",
    "roles": [
      {
        "_id": ObjectId("5a934e000102030405000001"),
        "approved": true,
        "id": 1,
        "name": "employeer"
      }
    ]
  },
  {
    "_id": ObjectId("5a934e000102030405000003"),
    "email": "admin@test.com",
    "password": "xxxxxxxx",
    "roles": [
      {
        "_id": ObjectId("5a934e000102030405000000"),
        "approved": true,
        "id": 0,
        "name": "administrator"
      },
      {
        "_id": ObjectId("5a934e000102030405000002"),
        "approved": true,
        "id": 2,
        "name": "freelancer"
      }
    ]
  }
]