Json MongoDB中两个集合的聚合

我有两个集合，我正在尝试查找它们。我不是100%确定如何接近它

系列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
    }
  }
 ]
}

{
 "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
   }
  }
]
}

集合2：条目

{
 "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
    }
  }
 ]
}

{
 "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
   }
  }
]
}

正如您所看到的，来自集合的employeeDetails.id与来自集合的条目的employeeId.id匹配

我正在尝试进行查找，以便将条目中的id与Employees中的用户名进行匹配，以便将名称与条目进行匹配，所需结果如下：

{
 "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
  }
 }
]
}

这将允许我获得组/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

我已经尝试了很多，唯一接近成功的方法是：

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

---

有人能给我建议/解决方案，告诉我实现目标的最佳方法吗？我选择了MongoDB，因为输入严重依赖于JSON输入。

您需要将

数据

数组本身的内容作为收集项。例如，您的

员工

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
    }
}

这样您就可以像这样轻松地

$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\u loaded

数组。

$group

阶段用于获取您想要保留的字段，并

$push

将它们放入一个名为

数据
的数组中。非常感谢这让我更亲近了。问题是，它给了我4个数组，全名，dateCreated，minutes和employeeId。它们只是列出的，但不匹配。我是否需要将上述内容包装为“匹配”或函数？链接到此处的示例：是否要匹配特定的用户名或employeeId？两个集合都具有employeeId，但只有Employee集合具有用户名。我希望能够在Entries集合中将Employee的用户名与employeeId相匹配。通过这种方式，我可以输出一个entires（分钟）列表，并让每个entires显示用户名。在entries集合中，您是否像employee集合中一样将数据数组项移动到集合本身？不，根据文章顶部的JSON示例，它们位于单独的集合中。