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Accepted
Marc Grue
Asked: 2023-11-06 03:39:53 +0800 CST

在过滤聚合中选择嵌套字段

  • 772

如何聚合、过滤嵌套数组和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
    }
  ]
}

..并且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，但如何仅返回嵌套学生数据中的name和字段？pets尝试以我能想到的所有方式来投影这些领域，但没有运气，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 ],
        }
      }
    }
  }}
])
mongodb

1 个回答

  1. Best Answer

    你可以添加一个$project舞台

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

    或者您可以使用$unset舞台删除您不需要的字段

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

    操场

    $addFields我认为后面加上一个会更干净$unset

    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
])

    操场

    • 1

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