在mongoose聚合期間保持原始欄位完整或通過管道傳輸它們-有解無憂

db.testA.aggregate([
  {
    "$lookup": {
      "from": "testB",
      "localField": "_id",
      "foreignField": "aId",
      "as": "join"
    }
  },
  {
    $project: {
      count: {
        $size: "$join"
      }
    }
  },
  {
    $set: {
      favorite: {
        $switch: {
          branches: [
            {
              case: {
                $gte: [
                  "$count",
                  1
                ]
              },
              then: true
            },
            {
              case: {
                $gte: [
                  "$count",
                  0
                ]
              },
              then: false
            },
            
          ],
          default: false
        }
      }
    }
  }
])

我有這個聚合功能。我想保留這些欄位，但我不確定它們為什么消失了。

db={
  "testA": [
    {
      "_id": ObjectId("60508eeb52a2ab598a013bb9"),
      "message": "hello",
      "field1": false,
      "field2": 1,
      "field3": "something"
    },
    {
      "_id": ObjectId("60508f2852a2ab598a013bba"),
      "message": "hello",
      "field1": false,
      "field2": 1,
      "field3": "something"
    },
    {
      "_id": ObjectId("60508f2852a2ab665a013bba"),
      "message": "hello",
      "field1": false,
      "field2": 1,
      "field3": "something"
    }
  ],
  "testB": [
    {
      "_id": ObjectId("60508f4152a2ab598a013bbb"),
      "aId": ObjectId("60508f2852a2ab598a013bba")
    },
    {
      "_id": ObjectId("60508f4552a2ab598a013bbc"),
      "aId": ObjectId("60508f2852a2ab598a013bba")
    },
    {
      "_id": ObjectId("605090a452a2ab598a013bbd"),
      "aId": ObjectId("60508eeb52a2ab598a013bb9")
    }
  ]
}

這是我的資料庫。本質上，我希望通過管道傳遞訊息 field1、field2、field3，以便最后得到 6 個欄位而不是 2 個欄位。

這是我得到的結果：

[
  {
    "_id": ObjectId("60508eeb52a2ab598a013bb9"),
    "count": 1,
    "favorite": true
  },
  {
    "_id": ObjectId("60508f2852a2ab598a013bba"),
    "count": 2,
    "favorite": true
  },
  {
    "_id": ObjectId("60508f2852a2ab665a013bba"),
    "count": 0,
    "favorite": false
  }
]

https://mongoplayground.net/p/KjquoKCUw4P

如果不清楚，這就是我想要的：

[
  {
    "_id": ObjectId("60508eeb52a2ab598a013bb9"),
    "count": 1,
    "favorite": true,
    "message": "hello",
    "field1": false,
    "field2": 1,
    "field3": "something"
  },
  {
    "_id": ObjectId("60508f2852a2ab598a013bba"),
    "count": 2,
    "favorite": true,
    "message": "hello",
    "field1": false,
    "field2": 1,
    "field3": "something"
  },
  {
    "_id": ObjectId("60508f2852a2ab665a013bba"),
    "count": 0,
    "favorite": false,
    "message": "hello",
    "field1": false,
    "field2": 1,
    "field3": "something"
  }
]

資料幾乎相同，但假設它們都不同。

uj5u.com熱心網友回復：

使用$addFields而不是 $project：

  {
    $addFields: {
      count: {
        $size: "$join"
      }
    }
  },

轉載請註明出處，本文鏈接：https://www.uj5u.com/qukuanlian/487248.html

標籤：mongodb 猫鼬聚合框架

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