嵌套模式/子文档对象中的猫鼬 findById() - 聚合

Posted 2023-03-11

【中文标题】嵌套模式/子文档对象中的猫鼬 findById() - 聚合

【英文标题】：Mongoose findById() in an object of nested schemas / subdocuments - aggregation

【发布时间】：2021-07-22 13:05:32

【问题描述】：

我有一个对象types。它包含多个模式。我需要通过 id 找到一个项目，但该项目可能在 exampleOne 或 exampleTwo 中（注意：将使用两个以上的模式）。

例如，在此查询"id: 608a5b290e635ece6828141e"：

  "_id": "608642db80a36336946620aa",
  "title": "titleHere",
  "types": 
    "exampleOne": [
      
        "_id": "6086430080a36336946620ab",
        "front": "front",
        "back": "back"
      ,
      
        "_id": "608a5b186ee1598ac9c222b4",
        "front": "front2",
        "back": "back2"
      
    ],
    "exampleTwo": [
      
        "_id": "608a5b290e635ece6828141e", // the queried document
        "normal": 
          "front": "2front",
          "back": "2back"
        ,
        "reversed": 
          "front": "2frontReversed",
          "back": "2backReversed"
        
      ,
      
        "_id": "608a5b31a3f9806de2537269",
        "normal": 
          "front": "2front2",
          "back": "2back2"
        ,
        "reversed": 
          "front": "2frontReversed2",
          "back": "2backReversed2"
        
      
    ]
  

应该返回：

  "_id": "608a5b290e635ece6828141e",
  "normal": 
    "front": "2front",
    "back": "2back"
  ,
  "reversed": 
    "front": "2frontReversed",
    "back": "2backReversed"
  
,

理想情况下，解决方案只需要一次搜索。我对此进行了一些研究，但无法弄清楚如何搜索 types 中的所有对象，而不为每个架构创建搜索并查看其中是否有任何返回结果。

如果需要，这是我的架构：

var MainSchema = new Schema (
  title:  type: String, required: true, maxlength: 255 ,
  types: 
    exampleOne: [exampleOneSchema],
    exampleTwo: [exampleTwoSchema],
  
);

var exampleOneSchema = new Schema(
    front: type: String, required: true,
    back: type: String, required: true,
);

var exampleTwoSchema= new Schema(
    normal: 
      front: type: String, required: true,
      back: type: String, required: true,
    ,
    reversed: 
      front: type: String, required: true,
      back: type: String, required: true,
    ,
);

感谢所有帮助！

谢谢，

Sour_Tooth

【参考方案1】：

演示 - https://mongoplayground.net/p/t5VYdkrL_nC

db.collection.aggregate([
  
    $match:  // filter the document so uniwnd and group have only 1 record to deal with
      $or: [
         "types.exampleOne._id": "608a5b290e635ece6828141e" ,
         "types.exampleTwo._id": "608a5b290e635ece6828141e" 
      ]
    
  ,
  
    $group: 
      _id: "$_id",
      docs:  $first:  "$concatArrays": [ "$types.exampleOne", "$types.exampleTwo" ]   // join both array into 1 element
    
  ,
   $unwind: "$docs" , //  break into individual documents
  
    $match:  // filter the records
     "docs._id": "608a5b290e635ece6828141e"
    
  ,
   $replaceRoot:  "newRoot": "$docs"   // set it to root
])

【讨论】：

我相信$match里面的$or是不必要的-mongoplayground.net/p/sFEZp6yQdK8

@Sour_Tooth 是的，我的错