将 mongodb 中的对象数组转换为字符串数组

Question

我正在查看以下文档。

下面将文档插入到集合中classes。

db.classes.insertMany( [
   { _id: 1, title: "Reading is ...", enrollmentlist: [ "giraffe2", "pandabear", "artie" ], days: ["M", "W", "F"] },
   { _id: 2, title: "But Writing ...", enrollmentlist: [ "giraffe1", "artie" ], days: ["T", "F"] }
] )

和members集合：

db.members.insertMany( [
   { _id: 1, name: "artie", joined: new Date("2016-05-01"), status: "A" },
   { _id: 2, name: "giraffe", joined: new Date("2017-05-01"), status: "D" },
   { _id: 3, name: "giraffe1", joined: new Date("2017-10-01"), status: "A" },
   { _id: 4, name: "panda", joined: new Date("2018-10-11"), status: "A" },
   { _id: 5, name: "pandabear", joined: new Date("2018-12-01"), status: "A" },
   { _id: 6, name: "giraffe2", joined: new Date("2018-12-01"), status: "D" }
] )

他们使用以下聚合来连接数组字段 上的两个集合enrollmentlist。

db.classes.aggregate( [
   {
      $lookup:
         {
            from: "members",
            localField: "enrollmentlist",
            foreignField: "name",
            as: "enrollee_info"
        }
   }
] )

返回以下内容：

{
   "_id" : 1,
   "title" : "Reading is ...",
   "enrollmentlist" : [ "giraffe2", "pandabear", "artie" ],
   "days" : [ "M", "W", "F" ],
   "enrollee_info" : [
      { "_id" : 1, "name" : "artie", "joined" : ISODate("2016-05-01T00:00:00Z"), "status" : "A" },
      { "_id" : 5, "name" : "pandabear", "joined" : ISODate("2018-12-01T00:00:00Z"), "status" : "A" },
      { "_id" : 6, "name" : "giraffe2", "joined" : ISODate("2018-12-01T00:00:00Z"), "status" : "D" }
   ]
}
{
   "_id" : 2,
   "title" : "But Writing ...",
   "enrollmentlist" : [ "giraffe1", "artie" ],
   "days" : [ "T", "F" ],
   "enrollee_info" : [
      { "_id" : 1, "name" : "artie", "joined" : ISODate("2016-05-01T00:00:00Z"), "status" : "A" },
      { "_id" : 3, "name" : "giraffe1", "joined" : ISODate("2017-10-01T00:00:00Z"), "status" : "A" }
   ]
}

如何将其简化enrolle_info为包含所有 s 的字符串数组name？

这就是我追求的结果：

{
   "_id" : 1,
   "title" : "Reading is ...",
   "enrollmentlist" : [ "giraffe2", "pandabear", "artie" ],
   "days" : [ "M", "W", "F" ],
   "enrollee_info" : [
      "artie",
      "pandabear"
      "giraffe2"
   ]
}
{
   "_id" : 2,
   "title" : "But Writing ...",
   "enrollmentlist" : [ "giraffe1", "artie" ],
   "days" : [ "T", "F" ],
   "enrollee_info" : [
      "artie",
      "giraffe1"
   ]
}

pipeline我还通过在操作中引入字段来研究使用多个联接$lookup。我可以使用 a$project来获取数组，{"name": "example"}但我不知道如何删除"name". 我尝试过使用{"$unwind": "$enrollee_info.name"}但那并没有给我我想要的。执行连接后，是否需要在聚合管道中引入另一个阶段？

Answer 1

看来我把这件事复杂化了。我通过执行以下操作达到了我想要的结果：

db.classes.aggregate( [
   {
      $lookup:
         {
            from: "members",
            localField: "enrollmentlist",
            foreignField: "name",
            as: "enrollee_info"
         }
   },
   {
      $project:
          {
              "_id": 1,
              "title": 1,
              "days": 1,
              "enrollee_names": "$enrollee_info.name"
          }
   } 
] )

结果：

[
    {
        "id": 1,
        "title": "Reading is ...",
        "days": [
            "M",
            "W",
            "F"
        ],
        "enrollee_names": [
            "artie",
            "pandabear",
            "giraffe2"
        ]
    },
    {
        "id": 2,
        "title": "But Writing ...",
        "days": [
            "T",
            "F"
        ],
        "enrollee_names": [
            "artie",
            "giraffe1"
        ]
    }
]