Mongoose 连接两个集合并仅从连接的集合中获取特定字段

我在加入猫鼬中的两个集合时遇到问题。我有两个集合，即：学生和考试。

学生模型：

{
  fullname: { type: String, required: true },
  email: { type: String, required: true },
}

考试模式：

{
  test: { type: String, required: false },
  top10: [
    type: {
      studentId: { type: String, required: true },
      score: { type: Number, required: false },
    }
  ]
}

现在，我想加入他们两个学生卡。结果应该是：

{
 "test": "Sample Test #1",
 "students": [
            {
                "studentId": "5f22ef443f17d8235332bbbe",
                "fullname": "John Smith",
                "score": 11
            },
            {
                "studentId": "5f281ad0838c6885856b6c01",
                "fullname": "Erlanie Jones",
                "score": 9
            },
            {
                "studentId": "5f64add93dc79c0d534a51d0",
                "fullname": "Krishna Kumar",
                "score": 5
            }
        ]
 }

我所做的是使用聚合：

 return await Exams.aggregate([
    {$lookup:
        {
            from: 'students',
            localField: 'top10.studentId',
            foreignField: '_id',
            as: 'students'
        }
    }
 ]);

但这个结果并不是我所希望的。有什么想法如何实现这一目标？我很乐意感谢任何帮助。谢谢！

你可以试试，

• $lookup with students收藏
• $project显示必填字段，$map迭代top10数组的循环并在内部使用$reduce获取学生的全名并使用以下命令与 top10 对象合并$mergeObjects

db.exams.aggregate([
  {
    $lookup: {
      from: "students",
      localField: "top10.studentId",
      foreignField: "_id",
      as: "students"
    }
  },
  {
    $project: {
      test: 1,
      students: {
        $map: {
          input: "$top10",
          as: "top10",
          in: {
            $mergeObjects: [
              "$$top10",
              {
                fullname: {
                  $reduce: {
                    input: "$students",
                    initialValue: 0,
                    in: {
                      $cond: [
                        { $eq: ["$$this._id", "$$top10.studentId"] },
                        "$$this.fullname",
                        "$$value"
                      ]
                    }
                  }
                }
              }
            ]
          }
        }
      }
    }
  }
])

操场 https://mongoplayground.net/p/_ioLktXuvgj

您可以使用的第二个选项$unwind before $lookup,

• $unwind解构top10 array
• $lookup with students收藏
• $addFields使用以下方法将学生数组转换为对象$arrayElemtAt
• $group通过 _id 并构建学生数组并推送所需字段

db.exams.aggregate([
  { $unwind: "$top10" },
  {
    $lookup: {
      from: "students",
      localField: "top10.studentId",
      foreignField: "_id",
      as: "students"
    }
  },
  { $addFields: { students: { $arrayElemAt: ["$students", 0] } } },
  {
    $group: {
      _id: "$_id",
      test: { $first: "$test" },
      students: {
        $push: {
          studentId: "$top10.studentId",
          score: "$top10.score",
          fullname: "$students.fullname"
        }
      }
    }
  }
])

操场 https://mongoplayground.net/p/rUF0jNiqkJM