MongoDB 使用关联的 collection 字段进行查找和排序的聚合正在减慢查询速度

Question

我的 mongodb 数据库中有两个 collection，如下所示：

employee_details with approximately 330000 documents which has department_id as a reference from departments collection

departments collections with 2 fields _id and dept_name

我已将索引添加到 collection 中，如下所示

db.departments.createIndex( { dept_name: 1 } )
db.employee_details.createIndex( { department_id: 1 } )
db.employee_details.createIndex( { employee_fname: 1 } )

我想通过加入两个 collection 来获取数据以在数据 table 上列出。但是当我尝试这样做时，我遇到了两个问题。

首先，当我在查找后添加排序时，查询运行需要很长时间，我添加了查找后排序，因为我需要对 collection 个部门的 dept_name 进行排序。查询如下

db.getCollection("employee_details").aggregate([
  {
    $lookup: {
      from: "departments",
      localField: "department_id",
      foreignField: "_id",
      as: "Department"
    }
  },
  { $unwind: { path: "$Department", preserveNullAndEmptyArrays: true } },
  { $sort: { "Department.dept_name": 1 } },
  { $limit: 30 }
]);

其次，当我在查找上方添加排序时，查询变得很快，但如果我使用 dept_name 或 department_id 进行排序，结果会给出错误的排序（排序对 [=15= 的字段工作正常] collection)。查询如下

db.getCollection("employee_details").aggregate([
  { $unwind: { path: "$Department", preserveNullAndEmptyArrays: true } },
  { $sort: { "Department.dept_name": 1 } },
  //{ $sort: { "department_id": 1 } }, // tried this also
  { $limit: 30 },
  {
    $lookup: {
      from: "departments",
      localField: "department_id",
      foreignField: "_id",
      as: "Department"
    }
  }
]);

有人可以提供优化的解决方案来获取数据以及从所有相关的 collection 中排序。提前谢谢你。

Answer 1

试试这个：

db.departments.aggregate([
  {
    $sort: {
      "dept_name": 1
    }
  },
  {
    $lookup: {
      from: "employee_details",
      localField: "_id",
      foreignField: "department_id",
      as: "Employee"
    }
  },
  {
    $unwind: "$Employee"
  },
  {
    $addFields: {
      tmp: {
        $mergeObjects: [
          {
            Department: "$$ROOT"
          },
          "$Employee"
        ]
      }
    }
  },
  {
    $project: {
      "tmp.Department.Employee": 0
    }
  },
  {
    $addFields: {
      "tmp.Department": [
        "$tmp.Department"
      ]
    }
  },
  {
    $replaceRoot: {
      newRoot: "$tmp"
    }
  },
  {
    $limit: 30
  }
])

MongoPlayground

Answer 2

您可以使用以下聚合：

db.getCollection("departments").aggregate([
    {
        $sort: {
            "dept_name": 1
        }
    },
    {
        $lookup: {
            from: "employee_details",
            localField: "_id",
            foreignField: "department_id",
            as: "employees"
        }
    },
    {
        $unwind: "$employees"
    },
    {
        $limit: 30
    },
    { $addFields: { "employees.department_name": "$dept_name" } },
    { $replaceRoot: { newRoot: "$employees" } }
])

这适用于 MongoDB 3.4 版，即使没有任何索引也足够快。

MongoDB 使用关联的 collection 字段进行查找和排序的聚合正在减慢查询速度

MongoDB aggregate with lookup and sort using associated collection fields is slowing down the query

sorting

lookup

limit

mongodb