Mongo 将多个文档聚合为一个

Question

我正在尝试通过 mongodb.collection.aggregate() 命令合并一些复杂的文档。

假设我要合并集合的文档的 x（在以下示例中：x=2）：

[
  {
    "_id": 1,
    "Data": {
      "children": {
        "1": {
          "name": "appear_only_in_first_doc",
          "cost": 1,
          "revenue": 4.5,
          "grandchildren": {
            "1t9dsqdqdvoj8pdppxjk": {
              "cost": 0,
              "revenue": 1.5
            }
          }
        },
        "2": {
          "name": "appear_in_both_docs",
          "cost": 2,
          "revenue": 7,
          "grandchildren": {
            "jesrdt5qwef2222dgt": {
              "cost": 1,
              "revenue": 3
            },
            "klh352hk5367kf": {
              "cost": 2,
              "revenue": 7
            }
          }
        }
      }
    }
  },
  {
    "_id": 2,
    "Data": {
      "children": {
        "2": {
          "name": "appear_in_both_docs___but_diff_name",
          "cost": 9,
          "revenue": 7,
          "grandchildren": {
            "aaaaaaaaa": {
              "cost": 3,
              "revenue": 2
            },
            "jesrdt5qwef2222dgt": {
              "cost": 6,
              "revenue": 5
            }
          }
        },
        "3": {
          "name": "appear_only_in_last_doc",
          "cost": 4,
          "revenue": 2,
          "grandchildren": {
            "cccccccccccc": {
              "cost": 4,
              "revenue": 2
            }
          }
        }
      }
    }
  }
]

挑战：

1. “children”和“grandchildren”键下的键是动态的，在编写查询时是未知的。
2. 如果子或孙子只出现在一个文档中（例如“1”、“3”、“1t9dsqdqdvoj8pdppxjk”、“klh352hk5367kf”、“aaaaaaaa”和“cccccccccccc”） - 它也应该出现在最终结果中。
3. 如果一个孩子出现在多个文档中（例如“2”和“jesrdt5qwef2222dgt”），它应该在最终结果中显示为一个。字段“成本”和“收入”应相加，最后一个“名称”字段应取。

我见过以下解决方案：

1. unionWith - 不相关，联合 2 个不同的集合。
2. merge - 不相关，不能对多次出现的字段值求和（取而代之的是最后一次）。
3. mergeObjects - 不相关，不能对多次出现的字段值求和（取而代之的是最后一次）。

最终的结果应该是这样的：

{
  "Data": {
    "children": {
      "1": {
        "name": "appear_only_in_first_doc",
        "cost": 1,
        "revenue": 4.5,
        "grandchildren": {
          "1t9dsqdqdvoj8pdppxjk": {
            "cost": 0,
            "revenue": 1.5
          }
        }
      },
      "2": {
        "name": "appear_in_both_docs___but_diff_name",
        "cost": 11,
        "revenue": 14,
        "grandchildren": {
          "aaaaaaaaa": {
            "cost": 3,
            "revenue": 2
          },
          "jesrdt5qwef2222dgt": {
            "cost": 7,
            "revenue": 8
          },
          "klh352hk5367kf": {
            "cost": 2,
            "revenue": 7
          }
        }
      },
      "3": {
        "name": "appear_only_in_last_doc",
        "cost": 4,
        "revenue": 2,
        "grandchildren": {
          "cccccccccccc": {
            "cost": 4,
            "revenue": 2
          }
        }
      }
    }
  }
}

Answer 1

这个过程有点长，可能会有一些简单的，我只是分享过程，

• $project 将children 转换为数组格式(k,v)
• $unwind解构children数组
• $group 按儿童键，对cost 和revenue 求和，最后使用$last 得到name
• $unwind解构grandchildren数组
• $addFields 将grandchildren 转换为数组格式(k,v)
• $unwind解构grandchildren数组

db.collection.aggregate([
  { $project: { "Data.children": { $objectToArray: "$Data.children" } } },
  { $unwind: "$Data.children" },
  {
    $group: {
      _id: "$Data.children.k",
      name: { $last: "$Data.children.v.name" },
      cost: { $sum: "$Data.children.v.cost" },
      revenue: { $sum: "$Data.children.v.revenue" },
      grandchildren: { $push: "$Data.children.v.grandchildren" }
    }
  },
  { $unwind: "$grandchildren" },
  { $addFields: { grandchildren: { $objectToArray: "$grandchildren" } } },
  { $unwind: "$grandchildren" },

• $group by children key 和 grandchildren key，计算孙子成本收入之和

  {
    $group: {
      _id: {
        ck: "$_id",
        gck: "$grandchildren.k"
      },
      cost: { $first: "$cost" },
      revenue: { $first: "$revenue" },
      name: { $first: "$name" },
      grandchildren_cost: { $sum: "$grandchildren.v.cost" },
      grandchildren_revenue: { $sum: "$grandchildren.v.revenue" }
    }
  },

• $group by children 键并重新构造 grandchildren 数组

  {
    $group: {
      _id: "$_id.ck",
      cost: { $first: "$cost" },
      revenue: { $first: "$revenue" },
      name: { $last: "$name" },
      grandchildren: {
        $push: {
          k: "$_id.gck",
          v: {
            cost: "$grandchildren_cost",
            revenue: "$grandchildren_revenue"
          }
        }
      }
    }
  },

• $group by null 并重新构造子数组并使用 $arrayToObject 将 grandchildren 转换为 (k,v) 数组中的对象

  {
    $group: {
      _id: null,
      children: {
        $push: {
          k: "$_id",
          v: {
            name: "$name",
            cost: "$cost",
            revenue: "$revenue",
            grandchildren: { $arrayToObject: "$grandchildren" }
          }
        }
      }
    }
  },

• $project 使用 $arrayToObject 将 children 转换为对象

  {
    $project: {
      _id: 0,
      "Data.children": { $arrayToObject: "$children" }
    }
  }
])

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