Mongo專案添加到陣列

我正在嘗試將一個專案添加到聚合中的陣列中，但似乎堅持了幾天。這是它的樣子：

db={
  Students: [
    {
      "id": "123",
      "title": "John",
      "marks_in_subjects": [
        {
          "marks": 90,
          "subject_id": "abc"
        },
        {
          "marks": 92,
          "subject_id": "def"
        }
      ]
    }
  ],
  Subjects: [
    {
      "id": "abc",
      "name": "math"
    },
    {
      "id": "def",
      "name": "physics"
    }
  ]
}

我對新欄位進行專案和連接標題，如下所示：

db.Students.aggregate([
  {
    $unwind: "$marks_in_subjects"
  },
  {
    "$lookup": {
      "from": "Subjects",
      "localField": "marks_in_subjects.subject_id",
      "foreignField": "id",
      "as": "subjects"
    }
  },
  {
    $unwind: "$subjects"
  },
  {
    $project: {
      _id: "$_id",
      title: "$title",
      titleAll: {
        $concat: [
          "$title",
          " > ",
          "$subjects.name",
          
        ]
      },
      
    }
  },
  
])

這是我得到的輸出：

[
  {
    "_id": ObjectId("5a934e000102030405000000"),
    "title": "John",
    "titleAll": "John \u003e math"
  },
  {
    "_id": ObjectId("5a934e000102030405000000"),
    "title": "John",
    "titleAll": "John \u003e physics"
  }
]

現在，我希望我的輸出在陣列中包含只有真正標題的第一項，然后列出所有這些，如下所示：

[
  {
    "_id": ObjectId("5a934e000102030405000000"),
    "title": "John",
    "titleAll": "John"
  },
  {
    "_id": ObjectId("5a934e000102030405000000"),
    "title": "John",
    "titleAll": "John \u003e math"
  },
  {
    "_id": ObjectId("5a934e000102030405000000"),
    "title": "John",
    "titleAll": "John \u003e physics"
  }
]

有沒有辦法實作它？我嘗試使用 $cond，但似乎不起作用。這是Mongo游樂場的鏈接

uj5u.com熱心網友回復：

你可以用更簡單的方式來做到這一點：

沒有必要一次$unwind又一次$group。缺少的步驟是將 連接title到titleAll陣列。

db.Students.aggregate([
  {
    $lookup: {
      from: "Subjects",
      localField: "marks_in_subjects.subject_id",
      foreignField: "id",
      as: "subjects"
    }
  },
  {
    $project: {subjects: "$subjects.name", title: 1}
  },
  {
    $addFields: {
      titleAll: {
        $map: {
          input: "$subjects",
          as: "item",
          in: {$concat: ["$title", " > ", "$$item"]}
        }
      }
    }
  },
  {
    $project: {
      title: "$title",
      titleAll: {"$concatArrays": ["$titleAll", ["$title"]]}
    }
  },
  {
    $unwind: "$titleAll"
  }
])

示例 MongoDB Playground

uj5u.com熱心網友回復：

1. $unwind
2. $lookup
3. $group- 分組_id。獲取title欄位和subjects陣列（第 4 階段需要此欄位）。
4. $setsubjects-通過添加第一個檔案和現有檔案來修改欄位subjects。
5. $unwind- 解構subjects欄位。
6. $project- 裝飾輸出檔案。

db.Students.aggregate([
  {
    $unwind: "$marks_in_subjects"
  },
  {
    "$lookup": {
      "from": "Subjects",
      "localField": "marks_in_subjects.subject_id",
      "foreignField": "id",
      "as": "subjects"
    }
  },
  {
    $group: {
      _id: "$id",
      title: {
        $first: "$title"
      },
      subjects: {
        $push: {
          $first: "$subjects"
        }
      }
    }
  },
  {
    $set: {
      subjects: {
        $concatArrays: [
          [
            {
              titleAll: "$title"
            }
          ],
          {
            $map: {
              input: "$subjects",
              in: {
                titleAll: {
                  $concat: [
                    "$title",
                    " > ",
                    "$$this.name"
                  ]
                }
              }
            }
          }
        ]
      }
    }
  },
  {
    $unwind: "$subjects"
  },
  {
    $project: {
      _id: 1,
      title: 1,
      titleAll: "$subjects.titleAll"
    }
  }
])

示例 Mongo Playground

標籤：mongodb

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