Mongoose将聚合到嵌套的doc Get count中，并将新值附加到查询结果中

Question

我有这个用户和帖子集合，其中包含嵌入式文档评论和回复：

{
 _id:34
 name:"dimer"
}

1. 职位

{
 _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，得到所有的帖子，之后我想给帖子、评论、回复添加新的键值，女巫会指定用户是否喜欢这个帖子，例如：喜欢： true注意：我已经准备好了经过身份验证的用户ID。我想检查用户ID是否存在于每个子树的喜欢文档中（帖子，评论，回复）

更具体地说，这是我预期的结果：

{
 _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最多的评论，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
  }
}

这就是我尝试的，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找到一个更干净的解决方案我的假解决方案




Answer 1

根据你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，您将得到如下响应-

{
"_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
    },
}}

如果你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" },
              }
      }
}])

以及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
}}