从 mongodb 中的三个集合中获取结果答案

【问题标题】：Get result from three collection in mongodb从 mongodb 中的三个集合中获取结果
【发布时间】：2018-06-20 11:03:02
【问题描述】：

我有三个集合包含如下数据，

客户集合

{ 
    "_id" : ObjectId("5a058e316803fafd127b23c9"), 
    "client_name" : "client A", 
    "client_status" : "A"
}
{ 
    "_id" : ObjectId("5a058e316803fafd127b23cb"), 
    "client_name" : "client B", 
    "client_status" : "A"
}

池集合

{ 
    "_id" : ObjectId("5a0e76f66803fa530a7b23d4"), 
    "pool_name" : "pool A", 
    "pool_status" : "A", 
    "client_id" : ObjectId("5a058e316803fafd127b23c9"), 
    "schools" : [
        ObjectId("5a0e742b6803faa6097b2462"), 
        ObjectId("5a0e742b6803faa6097b2464")
    ]
}
{ 
    "_id" : ObjectId("5a0e76f76803fa530a7b2402"), 
    "pool_name" : "pool B", 
    "pool_status" : "A", 
    "client_id" : ObjectId("5a058e316803fafd127b23c9"), 
    "schools" : [
        ObjectId("5a0e742b6803faa6097b2463")
    ]
}
{ 
    "_id" : ObjectId("5a0e76f76803fa530a7b23f6"), 
    "pool_name" : "pool C", 
    "pool_status" : "A", 
    "client_id" : ObjectId("5a058e316803fafd127b23cb"), 
    "schools" : [
        ObjectId("5a7aa1476803fa1f117b23d1")
    ]
}

学校收藏

{ 
    "_id" : ObjectId("5a0e742b6803faa6097b2462"), 
    "school_name" : "School A", 
    "school_status" : "A"
}
{ 
    "_id" : ObjectId("5a0e742b6803faa6097b2463"), 
    "school_name" : "School B", 
    "school_status" : "A"
}
{ 
    "_id" : ObjectId("5a0e742b6803faa6097b2464"), 
    "school_name" : "School C", 
    "school_status" : "A"
}
{ 
    "_id" : ObjectId("5a7aa1476803fa1f117b23d1"), 
    "school_name" : "School D", 
    "school_status" : "A"
}

从这个集合中，我需要为像下面这样的客户获取池的学校列表。

**Output**

{
    "_id" : ObjectId("5a058e316803fafd127b23c9"),
    "client_name" : "client A", 
    "client_status" : "A"
    "pools" : [
        {
            "_id" : ObjectId("5a0e76f66803fa530a7b23d4"), 
            "pool_name" : "pool A", 
            "pool_status" : "A",
            'schools' => [
                { 
                    "_id" : ObjectId("5a0e742b6803faa6097b2462"), 
                    "school_name" : "School A", 
                    "school_status" : "A"
                },
                { 
                    "_id" : ObjectId("5a0e742b6803faa6097b2464"), 
                    "school_name" : "School C", 
                    "school_status" : "A"
                },
            ]
        },
        {
            "_id" : ObjectId("5a0e76f76803fa530a7b2402"), 
            "pool_name" : "pool B", 
            "pool_status" : "A",
            "schools" : [
                { 
                    "_id" : ObjectId("5a0e742b6803faa6097b2463"), 
                    "school_name" : "School B", 
                    "school_status" : "A"
                }   
            ]
        }
    ]
},
{
    "_id" : ObjectId("5a058e316803fafd127b23cb"), 
    "client_name" : "client B",
    "client_status" : "A"
    "pools" : [
        {
            "_id" : ObjectId("5a0e76f76803fa530a7b23f6"), 
            "pool_name" : "pool C", 
            "pool_status" : "A",
            "schools" : [
                { 
                    "_id" : ObjectId("5a7aa1476803fa1f117b23d1"), 
                    "school_name" : "School D", 
                    "school_status" : "A"
                }
            ]
        }
    ]
}

【问题讨论】：

标签： mongodb collections aggregation-framework aggregate

【解决方案1】：

你可以试试下面的聚合

Mongodb 3.6 引入了嵌套的$lookup 管道......所以你不需要像上面那样使用另一个$lookup 阶段......你可以在这里使用嵌套的$lookup 管道

Client.aggregate([
  { "$match": { 'client_status': 'A' } },
  { "$lookup": {
    "from": Pool.collection.name,
    "let": { "client_id": "$_id" },
    "pipeline": [
      { "$match": { "$expr": { "$eq": [ "$client_id", "$$client_id" ] } } },
      { "$addFields": {
        "schools": {
          "$ifNull": ["$schools", []]
        }
      }}
      { "$lookup": {
        "from": Schools.collection.name,
        "let": { "schools": "$schools" },
        "pipeline": [
          { "$match": { "$expr": { "$in": [ "$_id", "$$schools" ] } } }
        ],
        "as": "schools"
      }}
    ],
    "as": "pools"
  }}
])

啰嗦的解释可以去$lookup multiple levels without $unwind?

【讨论】：

这会返回一个错误 - Mongo 服务器错误 (MongoCommandException): 命令失败，错误 40081: '$in 需要一个数组作为第二个参数，在服务器 localhost:27017 上找到：缺失'。完整的响应是： { "ok" : 0.0, "errmsg" : "$in 需要一个数组作为第二个参数，发现：缺失", "code" : NumberInt(40081), "codeName" : "Location40081" }
这是因为您的 schools 集合中的某处缺少您的 schools 密钥...尝试使用 schools: [] 更新您的集合
@Shijin 检查更新的答案...我在那边添加了$ifNull条件...现在它肯定会工作
如何获取每个客户的学校数量
@Shijin Hiii，如果您有新的疑问，请提出一个新问题...顺便说一句，您可以使用$size 运算符检查学生数组的长度