Mongoose 聚合到嵌套文档中获取计数并将新值附加到查询结果中答案

【问题标题】：Mongoose aggregate into a nested doc Get count and append new value to the query resultMongoose 聚合到嵌套文档中获取计数并将新值附加到查询结果中
【发布时间】：2021-08-04 15:18:06
【问题描述】：

我有这个用户和帖子集合以及嵌入式文档comments 和replies：

用户

{
 _id:34
 name:"dimer"
}

帖子

{
 _id:"1234",
 body:" hello ! I love mongodb, but its hard",
 likes:["34"],
 comments:{
    _id:"3453",
    body:"me I don't like mongodb i like sql "
    likes:["34"],
    replies:{
       _id:"2345",
       body:"both of them are great"
       likes:["34"],
    }
  }
}

问题是，我想聚合到帖子中并获取所有帖子，然后我想将新的键值附加到帖子、cmets、回复中，女巫会指定用户是否喜欢帖子：liked:true 注意：我已经准备好经过身份验证的用户 ID。我想检查每个子树（帖子、cmets、回复）上的likes 文档中是否存在用户 ID

为了更具体地了解这个问题，这是我预期的谦虚结果：

{
 _id:"1234",
 body:" hello ! I love mongodb, but its hard",
 liked:true,
 likesCount:1
 comments:{
     _id:"3453",
     body:"me I don't like mongodb i like sql "
     liked:true,
     likesCount:1
     replies:{
       _id:"2345",
       body:"both of them are great"
       liked:true,
       likesCount:1
    }
  }
}

注意：如果您可以帮助查询 cmets，对我来说没问题：预期结果：

{
 _id:"1234",
 body:" hello ! I love mongodb, but its hard",
 liked:true,
 likesCount:1
 comments:{
     _id:"3453",
     body:"me I don't like mongodb i like sql "
     liked:true,
     likesCount:1
  }
}

这是我尝试过的，根本不起作用：

const result = await PostModel.aggregate([
      {
        $project: {
          likesCount: { $size: "$likes" },
          commentsCount: { $size: "$comments" },
          liked: { $in: [ID(uid), "$likes"] },
          likes: 1,
          comments:{
             likesCount: { $size: "$likes" },
             liked: { $in: [ID(uid), "$likes"] },
            ......
         } 
     }
}

... 注意：我想使用 $group 和 $unwind 来找到一个更清洁的解决方案我的假solution

【问题讨论】：

标签： javascript arrays reactjs mongodb mongoose

【解决方案1】：

根据您的第一个预期结果，这是一个解决方案

const result = await PostModel.aggregate([
{
      $project: {
          _id: 1,
          body: 1,
          liked: { $in: [ID(uid), "$likes"] },
          likesCount: { $size: "$likes" },
          comments: {
                _id: 1,
                body:1,
                liked: { $in: [ID(uid), "$comment.likes"] },
                likesCount: { $size: "$comment.location.likes" },
               
      }
}]);

通过这个查询，你会得到这样的响应-

{
"_id" : "1234",
"body" : " hello ! I love mongodb, but its hard",
"liked" : true,
"likesCount" : 1
"comments" : {
    "_id" : 3453",
    "body" : "me I don't like mongodb i like sql ",
    "liked" : true,
    "likesCount" : 1
    "replies" : {
        "_id" : "2345",
        "body" : "both of them are great",
        "liked" : true,
        "likesCount" : 1
    },
}}

如果您想要第二个预期结果，那么这是一个解决方案 -

const result = await PostModel.aggregate([
{
      $project: {
          _id: 1,
          body: 1,
          liked: { $in: [ID(uid), "$likes"] },
          likesCount: { $size: "$likes" },
          comments: {
                _id: 1,
                body:1,
                liked: { $in: [ID(uid), "$likes"] },
                likesCount: { $size: "$likes" },
              }
      }
}])

由此产生的预期结果 -

{
"_id" : "1234",
"body" : " hello ! I love mongodb, but its hard",
"liked" : true,
"likesCount" : 1
"comments" : {
    "_id" : "3453",
    "body" : "me I don't like mongodb i like sql ",
    "liked" : true,
    "likesCount" : 1
}}

【讨论】：