Mongodb Select 筛选聚合中的嵌套字段

发布于11月06日

如何聚合、筛选嵌套数组并投影选定的嵌套字段？

我有以下数据:

{
  "teacher": "Ben",
  "students": [
    {
      "name": "Liz",
      "age": 8,
      "pets": 1
    },
    {
      "name": "Jon",
      "age": 9,
      "pets": 3
    },
    {
      "name": "Eva",
      "age": 9,
      "pets": 2
    }
  ]
}

..并想要查询以获得此结果:

{
  "teacher": "Ben",
  "students": [
    {
      "name": "Jon",
      "pets": 3
    },
    {
      "name": "Eva",
      "pets": 2
    }
  ]
}

我可以按某个年龄进行过滤，但如何才能只返回嵌套的学生数据中的name和pets个字段呢？我试图以所有我能想到的方式投射这些油田，但运气不佳，所以我猜我可能找错了地方.

db.classes.aggregate([
  { "$match": {} },
  { "$project": {
    "_id": 0,
    "teacher": 1,
    "students": {
      "$filter": {
        input: "$students",
        cond: {
          "$eq": [ "$$this.age", 9 ],
        }
      }
    }
  }}
])

推荐答案

你可以加一个$project级的stage

 //previous stages 
 {
    $project: {
      "students.age": 0
    }
  }

或者，您可以使用$unset阶段删除不需要的字段

db.collection.aggregate([
  { $match: {} },
  {
    $project: {
      _id: 0,
      teacher: 1,
      students: {
        $filter: {
          input: "$students",
          cond: { $eq: [ "$$this.age", 9 ] }
        }
      }
    }
  },
  { $unset: "students.age" }
])

playground个

我觉得如果先考$addFields分，再考$unset分，那就更干净了

db.collection.aggregate([
  { $match: {} },
  {
    $addFields: {
      students: {  //fields you just want to add, here overwrites students
        $filter: {
          input: "$students",
          cond: { $eq: [ "$$this.age", 9 ] }
        }
      }
    }
  },
  { $unset: [ "students.age", "_id" ] } //remove just the ones not needed
])

playground个