如何对 MongoDB 中数组内的字段值进行分组？答案

【问题标题】：How to groupby on a field value that is inside an array in MongoDB?如何对 MongoDB 中数组内的字段值进行分组？
【发布时间】：2021-10-06 12:56:55
【问题描述】：

我有使用 MongoDB 的学生评分系统。我在 MongoDB 中有以下文档：

我如何获取特定subject 获得grade "A","B","C","D" 的学生人数。每个主题的“_id”保持不变。

{
  _id: "60b223541338467beaf3ae0d",
  studentName: "John Doe"
  studentGradingDetails: [
    {
      _id: "60b21e47e5462929cab27a98",
      term: "semester-1",
      subject: "chemistry",
      grade: "A",
      createdAt: "2021-05-29T10:58:15.113Z",
    },
    {
      _id: "60b21e47e5462929cab27a99",
      term: "semester-2",
      subject: "computer_science",
      grade: "B",
      createdAt: "2021-05-29T10:58:15.113Z",
    },
  ],
  createdAt: "2021-05-29T11:19:48.770Z",
}
{
  _id: "60b223541338467beaf3ae0e",
  studentName: "Will Smith"
  studentGradingDetails: [
    {
      _id: "60b21e47e5462929cab27a98",
      term: "semester-1",
      subject: "chemistry",
      grade: "D",
      createdAt: "2021-05-29T10:58:15.113Z",
    },
    {
      _id: "60b21e47e5462929cab27a99",
      term: "semester-2",
      subject: "computer_science",
      grade: "A",
      createdAt: "2021-05-29T10:58:15.113Z",
    },
  ],
  createdAt: "2021-05-29T11:19:48.770Z",
}

这是我尝试并坚持的，不知道接下来的步骤是什么！？


await db.collection("studentsemestergrades")
        .aggregate([
            { $unwind: "$studentGradingDetails" },
            {
                $group: {
                    _id: "$studentGradingDetails._id",
                    
                },
            },
        ])
        .toArray();

预期输出：

每个科目得分不同的学生分布(_id)

{
  "_id" : "60b21e47e5462929cab27a98",
  "A" : 12,
  "B" : 20,
  "C" : 8,
  "D" : 2
},
{
  "_id" : "60b21e47e5462929cab27a99",
  "A" : 5,
  "B" : 2,
  "C" : 8,
  "D" : 12
}

【问题讨论】：

标签： node.js mongodb aggregate-functions

【解决方案1】：

$unwind 解构 studentGradingDetails 数组
$group by _id 和 grade 并获得总数
$group 仅由_id 和grade 和count 构造对象的键值对数组
$arrayToObject将上面的键值对数组转换为对象
$mergeObjects 合并以上转换后的对象和 _id 字段，如果需要，您可以添加更多字段
$replaceRoot 将上面的合并对象替换为根

await db.collection("studentsemestergrades").aggregate([
  { $unwind: "$studentGradingDetails" },
  {
    $group: {
      _id: {
        _id: "$studentGradingDetails._id",
        grade: "$studentGradingDetails.grade"
      },
      count: { $sum: 1 }
    }
  },
  {
    $group: {
      _id: "$_id._id",
      grades: {
        $push: {
          k: "$_id.grade",
          v: "$count"
        }
      }
    }
  },
  {
    $replaceRoot: {
      newRoot: {
        $mergeObjects: [
          { _id: "$_id" },
          { $arrayToObject: "$grades" }
        ]
      }
    }
  }
])

Playground

【讨论】：