MongoDb聚合，用于根据相同集合的所有文档中数组对象中存在的ID过滤列表 [英] MongoDb aggregation for filtering list based on ids present in object of array from all document of same collection

问题描述

在我的mongodb中，我的数据如下所示:

In my mongodb I have the data as shown below:

{
  "studentId": "a1",
  "name":"John Doe"
  "studentsReffered": [    
    {
      "course": "BTECH",
      "students": [
        {
          "studentId": "a2"
        },
        {
          "studentId": "a3"
        }
      ]
    },
    {
      "course": "MTECH",
      "students": [        
        {
          "studentId": "a4"
        },
        {
          "studentId": "a5"
        }
      ]
    }
  ]
},
{
  "studentId": "a2",
  "name":"Joseph"
  "studentsReffered": [    
    {
      "course": "BTECH",
      "students": [
        {
          "studentId": "a6"
        }
      ]
    }
  ]
}

以上JSON包含MongoDB集合中的文档.每个文档都包含学生的所有详细信息以及推荐详细信息摘要.即，对于每个学生，都有一个fieldsReferred字段，其中包含该学生推荐的所有学生的ID.

Above JSON contains documents in collection of MongoDB. Each document contains all details of student along with referral detail summary. i.e. for every student there is a field studentsReferred which contain ids of all students which are referred by the student.

我想单独显示所有学生的详细信息，以及在检索单个学生时被推荐的学生的姓名.如下

I want to show all details of student alone with the name of students which are reffered while retrieving the single student. As below

{
  "studentId": "a1",
  "name":"John Doe"
  "studentsReffered": [    
    {
      "course": "BTECH",
      "students": [
        {
          "studentId": "a2",
          "name":"Joseph"
        },
        {
          "studentId": "a3",
          "name":"Lorem Ipsum"
        }
      ]
    },
    {
      "course": "MTECH",
      "students": [        
        {
          "studentId": "a4",
          "name":"Lorem Ipsum"
        },
        {
          "studentId": "a5",
          "name":"Lorem Ipsum"
        }
      ]
    }
  ]
}

我尝试使用mongodb聚合解决此问题.但不幸的是，我无法为此编写查询.因此，我们可以使用聚合来实现上述方案.

I have tried to use mongodb aggregation for this problem. But unfortunately I am not able write query for that. So can we achieve the above scenario using aggregation.

推荐答案

您可以尝试

• $ facet 创建2个数组， users 用户详细信息名称和studentId，第二个所有用户详细信息在 allUsers
中
• $ project 迭代循环
  • $ map 输入为allUsers数组
  • $ map 输入为studentsReffered数组
  • $ map 输入为学生数组
  • $ reduce 在条件匹配时从 users 数组中获取学生的数据
  • $facet to create 2 arrays, users user details name and studentId, second all users details in allUsers
  • $project iterate loop
    • $map input as allUsers array
    • $map input as studentsReffered array
    • $map input as students array
    • $reduce to get data of the student from users array when condition match

    db.collection.aggregate([
      {
        $facet: {
          users: [
            {
              $project: {
                studentId: 1,
                name: 1
                // add fields as you want it will automatically reflect in join
              }
            }
          ],
          allUsers: []
        }
      },
      {
        $project: {
          allUsers: {
            $map: {
              input: "$allUsers",
              in: {
                $mergeObjects: [
                  "$$this",
                  {
                    studentsReffered: {
                      $map: {
                        input: "$$this.studentsReffered",
                        in: {
                          $mergeObjects: [
                            "$$this",
                            {
                              students: {
                                $map: {
                                  input: "$$this.students",
                                  as: "s",
                                  in: {
                                    $reduce: {
                                      input: "$users",
                                      initialValue: { studentId: "$$s.studentId" },
                                      in: {
                                        $cond: [
                                          { $eq: ["$$this.studentId", "$$s.studentId"] },
                                          "$$this",
                                          "$$value"
                                        ]
                                      }
                                    }
                                  }
                                }
                              }
                            }
                          ]
                        }
                      }
                    }
                  }
                ]
              }
            }
          }
        }
      },
      { $unwind: "$allUsers" },
      { $replaceWith: "$allUsers" }
    ])
    

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