如何在 MongoDB 中的嵌套嵌入 object 上聚合

Question

{
  "_id": 1,
  "title": {
    header1:{ "name":"A" , age:10, "active":true},
    header2:{ "name":"B" , age:15, "active":false},
    header3:{ "name":"C" , age:20, "active":true},
    header4:{ "name":"D" , age:30, "active":true},
    header5:{ "name":"E" , age:35, "active":false},
    header6:{ "name":"F" , age:40, "active":true},
    header7:{ "name":"G" , age:45, "active":false},
  }
}

如何只显示那些 headers 其活动值为 false 的？

Answer 1

1. $project

   1.1 通过$objectToArray for headers 字段将键值对转换为数组。 1.2 $filter 与 $$this.v.active 即 false.

2. $project

   2.1 通过$arrayToObject将headers转换为键值对并赋值给title字段

db.collection.aggregate([
  {
    $project: {
      _id: 1,
      headers: {
        $filter: {
          "input": {
            $objectToArray: "$title"
          },
          "cond": {
            "$eq": [
              "$$this.v.active",
              false
            ]
          }
        }
      }
    }
  },
  {
    $project: {
      _id: 1,
      title: {
        "$arrayToObject": "$headers"
      }
    }
  }
])

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