Question

我正在尝试使用聚合框架创建查询，但无法获得所需的结果。我有一个代理商集合，每个代理商都有一个客户列表，每个客户都有一个成员列表，结构如下：

[
{
  "userID" : "xxx",
  "userType" : "RESELLER",
  "clients" : [
     {
        "userID" : "xxx",
        "userType" : "CLIENT",
        "members" : [
           {
              "userID" : "xxx",
              "userType" : "MEMBER"
           },
           {
              "userID" : "xxx",
              "userType" : "MEMBER"
           }
        ]
     },
{
        "userID" : "xxx",
        "userType" : "CLIENT",
        "members" : [
           {
              "userID" : "xxx",
              "userType" : "MEMBER"
           },
           {
              "userID" : "xxx",
              "userType" : "MEMBER"
           }
        ]
     }
   ]
},
{
  "userID" : "xxx",
  "userType" : "RESELLER",
  "clients" : [
     {
        "userID" : "xxx",
        "userType" : "CLIENT",
        "members" : [
           {
              "userID" : "xxx",
              "userType" : "MEMBER"
           },
           {
              "userID" : "xxx",
              "userType" : "MEMBER"
           }
        ]
     },
{
        "userID" : "xxx",
        "userType" : "CLIENT",
        "members" : [
           {
              "userID" : "xxx",
              "userType" : "MEMBER"
           },
           {
              "userID" : "xxx",
              "userType" : "MEMBER"
           }
        ]
     }
   ]
}
]

我想要得到的结果是：

[
   {
      "userID" : "xxx",
      "userType" : "RESELLER"
   },
   {
      "userID" : "xxx",
      "userType" : "RESELLER"
   },
   {
      "userID" : "xxx",
      "userType" : "CLIENT"
   },
   {
      "userID" : "xxx",
      "userType" : "CLIENT"
   },
   {
      "userID" : "xxx",
      "userType" : "CLIENT"
   },
   {
      "userID" : "xxx",
      "userType" : "CLIENT"
   },
   {
      "userID" : "xxx",
      "userType" : "MEMBER"
   },
   {
      "userID" : "xxx",
      "userType" : "MEMBER"
   },
   {
      "userID" : "xxx",
      "userType" : "MEMBER"
   },
   {
      "userID" : "xxx",
      "userType" : "MEMBER"
   },
   {
      "userID" : "xxx",
      "userType" : "MEMBER"
   }
]

我做了很多尝试，但没有得到这个结果。我最接近的解决方案是：

db.resellers.aggregate([
{
    $unwind: "$clients"
},
{
    $project: {
        _id : 0,
        teamMembers : "$clients.members"
    }
},
{
    $unwind: "$members"
},
{
    $project: {
        _id : 0,
        userID : "$members.userID",
        type : "$members.type"
    }
}
]).pretty()

此解决方案仅返回成员列表，那么我需要做些什么来获得包含所有经销商，客户和成员的列表？

Answer 1

您可以将$reduce与$concatArrays结合使用来展平数据结构，然后将$unwind与$replaceRoot一起运行以获取每个文档的单个成员：

jump

Mongo Playground

Answer 2

好吧，我在此向您介绍$project阶段中的所有内容。享受吧！

[
    {
        "$project": {
            "members": {
                "$reduce": {
                    "input": {
                        "$map": {
                            "input": "$clients",
                            "in": {
                                "$concatArrays": [
                                    [
                                        {
                                            "userID": "$userID",
                                            "userType": "$userType"
                                        },
                                        {
                                            "userID": "$$this.userID",
                                            "userType": "$$this.userType"
                                        }
                                    ],
                                    "$$this.members"
                                ]
                            }
                        }
                    },
                    "initialValue": [

                    ],
                    "in": {
                        "$concatArrays": [
                            "$$this",
                            "$$value"
                        ]
                    }
                }
            }
        }
    }
]

Playground

合并文档及其嵌套数组和嵌套数组

2 个答案: