mongodb-将嵌套对象数组合并为一个，保留所有数据

如何解决mongodb-将嵌套对象数组合并为一个，保留所有数据

我在mongodb 4.4.0中有2个集合。 我有一个包含所有视频和一些相关字段的视频集合，还有一个称为用户的集合，我在其中存储所有用户以及他们已完成的视频列表，包括他们在单个视频上放置的注释。 现在，我必须获取所有视频的列表，并结合用户集合中包含的视频字段，仅在明显存在对应关系的情况下

视频

{
  "_id": ObjectId("aaaaaaaaaaaaaaaaaaaaaa01"),"title": "Video title 1","duration" : 120,"author" : "John Doe"
},{
  "_id": ObjectId("aaaaaaaaaaaaaaaaaaaaaa02"),"title": "Video title 2","duration" : 180,"author" : "Maria Hernandez"
},{
  "_id": ObjectId("aaaaaaaaaaaaaaaaaaaaaa03"),"title": "Video title 3","duration" : 75,"author" : "Henry Ford"
}

用户（仅一个用于演示）

{
"_id": ObjectId("bbbbbbbbbbbbbbbbbbbbbb01"),"fullname": "James Smith","videos":
 [
  {
    "video_id": ObjectId("aaaaaaaaaaaaaaaaaaaaaa01"),"views": 12,"completed": true,"notes": "very fun video","tags": ["fun","best"]
  },{
    "video_id": ObjectId("aaaaaaaaaaaaaaaaaaaaaa02"),"views": 64,"completed": false,"notes": "very difficult to understand","tags": ["hard","training","sport"]
  }
 ]
}

期望的结果合并对象，假设我正在为用户bbbbbbbbbbbbbbbbbbbbbbbbbb01进行过滤：

{
    "_id": ObjectId("bbbbbbbbbbbbbbbbbbbbbb01"),"videos":
     [
      {
        "video_id": ObjectId("aaaaaaaaaaaaaaaaaaaaaa01"),"author" : "John Doe"
        "views": 12,"best"]
      },{
        "video_id": ObjectId("aaaaaaaaaaaaaaaaaaaaaa02"),"author" : "Maria Hernandez"
        "views": 64,"sport"]
      },{
        "video_id": ObjectId("aaaaaaaaaaaaaaaaaaaaaa03"),"author" : "Henry Ford"
      }
     ]
    }

例如，我在Google和SO上进行了大量搜索： MongoDB - $lookup in complex nested array或 Merge two array objects together in Mongodb from $Lookup

我正在尝试使用$查找和$ mergeObjects，但是我无法获得想要的结果。

解决方法

您可以尝试

• $group设为null，并制作users数组

db.users.aggregate([
  {
    $group: {
      _id: null,users: { $push: "$$ROOT" }
    }
  },

• $lookup具有视频收藏夹，并以videos数组的形式获取所有视频

  {
    $lookup: {
      from: "videos",pipeline: [],as: "videos"
    }
  },

• $project更新users.videos数组
• $map迭代用户数组的循环，合并来自内部地图的当前对象和视频数组，
• $map迭代来自视频集合的视频数组循环
• $reduce迭代users.videos循环，如果video_id匹配则匹配条件，然后返回视频对象并与视频合并，否则返回空白对象。

  {
    $project: {
      users: {
        $map: {
          input: "$users",as: "user",in: {
            $mergeObjects: [
              "$$user",{
                videos: {
                  $map: {
                    input: "$videos",as: "video",in: {
                      $mergeObjects: [
                        "$$video",{
                          $reduce: {
                            input: "$$user.videos",initialValue: {},in: {
                              $cond: [
                                { $eq: ["$$video._id","$$this.video_id"] },"$$this","$$value"
                              ]
                            }
                          }
                        }
                      ]
                    }
                  }
                }
              }
            ]
          }
        }
      }
    }
  },

• $unwind解构用户数组
• $replaceRoot替换root用户对象

  { $unwind: "$users" },{ $replaceRoot: { newRoot: "$users" } }
])

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