MongoDB中两个集合的聚合

Question

我有两个集合，我正在尝试对其进行某种查找。我不是 100% 确定如何处理它。

集合 1：员工

{
 "data": [
  {
     "full name": "Keith Richards",
     "age": 21,
     "userName": "keith1@keith.com",
     "employeeDetails": {
        "id": 102522
    }
  },
  {
    "full name": "Jim Morrison",
     "age": 27,
     "userName": "jim@jim.com",
     "employeeDetails": {
        "id": 135522
    }
  }
 ]
}

收藏 2：参赛作品

{
 "data": [
  {
     "dateCreated": "02-04-2016",
     "billable": true,
     "minutes": 150,
     "employeeId": {
        "id": 102522
        }
  },
  {
     "dateCreated": "03-04-2016",
     "billable": true,
     "minutes": 250,
     "employeeId": {
        "id": 102522
   }
  },
  {
     "dateCreated": "04-04-2016",
     "billable": true,
     "minutes": 20,
     "employeeId": {
        "id": 135522
   }
  }
]
}

您可以看到 Employees 集合中的 employeeDetails.id 与 Entries 中的 employeeId.id 匹配> 收藏。

我正在尝试进行查找，以便将条目中的 id 与员工中的用户名匹配，这样我就可以将名称与条目匹配，所需的结果如下：

{
 "data": [
  {
     "full name": "Keith Richards",
     "dateCreated": "02-04-2016",
     "minutes": 150,
     "employeeId": {
        "id": 102522
        }
  },
  {
     "full name": "Keith Richards",
     "dateCreated": "03-04-2016",
     "minutes": 250,
     "employeeId": {
        "id": 102522
        }
  },
  {
     "full name": "Jim Morrison",
     "dateCreated": "04-04-2016",
     "minutes": 20,
     "employeeId": {
        "id": 135522
  }
 }
]
}

这将允许我得到一个组/总和的最终结果：

• 全名：基思·理查兹
• 分钟：400
• 全名：吉姆·莫里森
• 分钟：20

我尝试了一堆，唯一接近工作的是以下：

db.getCollection('entries')
.aggregate(
  [
    {
      "$lookup": {
        "from": "employees", 
        "localField": "trafficEmployeeId.id", 
        "foreignField": "employeeDetails.id", 
        "as": "employees_loaded"
      }
    }
  ]
);

这基本上给了我一个包含两个不同数组的集合。

谁能给我建议/解决方案，以最佳方式实现我想要实现的目标？我选择了 MongoDB，因为输入严重依赖 JSON 输入。

Answer 1

您需要将data 数组本身的内容包含为集合项。例如，您的 employees 集合应如下所示：

{
    "full name": "Keith Richards",
    "age": 21,
    "userName": "keith1@keith.com",
    "employeeDetails": {
        "id": 102522
    }
}, {
    "full name": "Jim Morrison",
    "age": 27,
    "userName": "jim@jim.com",
    "employeeDetails": {
        "id": 135522
    }
}

这样你就可以像这样轻松$lookup：

db.entries.aggregate(
    [
        { $unwind: "$data" }, {
            $lookup: {
                "from": "employees",
                "localField": "data.employeeId.id",
                "foreignField": "employeeDetails.id",
                "as": "employees_loaded"
            }
        }, {
            $unwind: "$employees_loaded"
        }, {
            $group: {
                "_id": 1,
                "data": {
                    $push: {
                        "full name": "$employees_loaded.full name",
                        "dateCreated": "$data.dateCreated",
                        "minutes": "$data.minutes",
                        "employeeId": "$data.employeeId"
                    }
                }
            }
        }
    ]
);

$unwind 用于删除从$lookup 生成的employees_loaded 数组。 $group 阶段用于获取您想要保留的字段并将$push 它们放入一个名为data 的数组中