MongoDb 聚合与数组内的数组可能

Question

我正在努力寻找一些使用 mongo 聚合框架来处理具有项目数组的文档的示例，其中每个项目还具有其他对象的数组（包含数组的数组）

在下面的示例文档中，我真正想要的是一个示例，它对文档中所有案例的结果数组中的 itemValue 求和，并跨越集合，其中 result.decision 被“接受”并按文档 locationCode

但是，即使是找到所有文档的示例，其中 result.decision 被“接受”以显示或汇总 itemValue 的相同内容也会有所帮助

非常感谢

{
"_id": "333212",
"data": {
    "locationCode": "UK-555-5566",
    "mode": "retail",
    "caseHandler": "A N Other",
    "cases": [{
            "caseId": "CSE525666",
            "items": [{
                    "id": "333212-CSE525666-1",
                    "type": "hardware",
                    "subType": "print cartridge",
                    "targetDate": "2020-06-15",
                    "itemDetail": {
                        "description": "acme print cartridge",
                        "quantity": 2,
                        "weight": "1.5"
                    },
                    "result": {
                        "decision": "rejected",
                        "decisionDate": "2019-02-02"
                    },
                    "isPriority": true
                },
                {
                    "id": "333212-CSE525666-2",
                    "type": "Stationery",
                    "subType": "other",
                    "targetDate": "2020-06-15",
                    "itemDetail": {
                        "description": "staples box",
                        "quantity": 3,
                        "weight": "1.66"
                    },
                    "result": {
                        "decision": "accepted",
                        "decisionDate": "2020-03-03",
                        "itemValue": "23.01"
                    },
                    "isPriority": true
                }
            ]
        },
        {
            "caseId": "CSE885655",
            "items": [{
                    "id": "333212-CSE885655-1",
                    "type": "marine goods",
                    "subType": "fish food",
                    "targetDate": "2020-06-04",
                    "itemDetail": {
                        "description": "fish bait",
                        "quantity": 5,
                        "weight": "0.65"
                    },
                    "result": {
                        "decision": "accepted",
                        "decisionDate": "2020-03-02"
                    },
                    "isPriority": false
                },
                {
                    "id": "333212-CSE885655-4",
                    "type": "tobacco products",
                    "subType": "cigarettes",
                    "deadlineDate": "2020-06-15",
                    "itemDetail": {
                        "description": "rolling tobbaco",
                        "quantity": 42,
                        "weight": "2.25"
                    },
                    "result": {
                        "decision": "accepted",
                        "decisionDate": "2020-02-02",
                        "itemValue": "48.15"
                    },
                    "isPriority": true
                }
            ]
        }
    ]
},
"state": "open"

}

Answer 1

您可能正在寻找$unwind。它在文档中获取一个数组，并为每个数组成员创建一个单独的文档。

{ foos: [1, 2] } -> { foos: 1 }, { foos: 2}

这样您就可以创建一个平面文档结构并像往常一样进行匹配和分组。

db.collection.aggregate([
  {
    $unwind: "$data.cases"
  },
  {
    $unwind: "$data.cases.items"
  },
  {
    $match: {
      "data.cases.items.result.decision": "accepted"
    }
  },
  {
    $group: {
      _id: "$data.locationCode",
      value: {
        $sum: {
          $toDecimal: "$data.cases.items.result.itemValue"
        }
      }
    }
  },
  {
    $project: {
      _id: 0,
      locationCode: "$_id",
      value: "$value"
    }
  }
])

https://mongoplayground.net/p/Xr2WfFyPZS3

Answer 2

替代解决方案...

我们按data.locationCode 分组并将所有符合此条件的项目相加：
cases[*].items[*].result.decision" == "accepted"

---

db.collection.aggregate([
  {
    $group: {
      _id: "$data.locationCode",
      itemValue: {
        $sum: {
          $reduce: {
            input: "$data.cases",
            initialValue: 0,
            in: {
              $sum: {
                $concatArrays: [
                  [ "$$value" ],
                  {
                    $map: {
                      input: {
                        $filter: {
                          input: "$$this.items",
                          as: "f",
                          cond: {
                            $eq: [ "$$f.result.decision", "accepted" ]
                          }
                        }
                      },
                      as: "item",
                      in: {
                        $toDouble: {
                          $ifNull: [ "$$item.result.itemValue", 0 ]
                        }
                      }
                    }
                  }
                ]
              }
            }
          }
        }
      }
    }
  }
])

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