Node.js 仅当所需文档是最后一个文档时才更新MongoDB，否则插入

虽然我认为将您的现有工作包装在transactions中应该是涉及最少更改的选项，但您可以 Select 为所有工作利用聚合管道.

1. $match以判断现有数据
2. $sort+$limit: 1获得最后一项记录
3. $set+$add以递增计数器
4. $setUnion与您的默认数据"插入模板"
5. $setWindowFields to compute $rank for the user_id partition
   • sortBy timestamp到计算机rank
   • 如果有来自步骤2，3的现有数据，它将具有等级:1，因为它的时间戳将小于$$NOW当前时间戳"插入模板"
   • 如果不存在现有数据，则"插入模板"的排名为:1
6. $match用于仅提取rank: 1数据，$unset用于清理rank字段
7. $merge以插入回集合中

db.notifications.aggregate([
  {
    "$match": {
      // your input user_id and message to upsert here
      "user_id": 1,
      "message": "update case - count should be incremented"
    }
  },
  {
    "$sort": {
      "timestamp": -1
    }
  },
  {
    "$limit": 1
  },
  {
    "$set": {
      "count": {
        "$add": [
          "$count",
          1
        ]
      }
    }
  },
  {
    "$unionWith": {
      "coll": "notifications",
      "pipeline": [
        {
          "$documents": [
            // document to insert if not found
            {
              "user_id": 1,
              "message": "msg to insert",
              "count": 1,
              "timestamp": "$$NOW"
            }
          ]
        }
      ]
    }
  },
  {
    "$setWindowFields": {
      "partitionBy": "$user_id",
      "sortBy": {
        "timestamp": 1
      },
      "output": {
        "rank": {
          "$rank": {}
        }
      }
    }
  },
  {
    "$match": {
      "rank": 1
    }
  },
  {
    "$unset": "rank"
  },
  {
    "$merge": {
      "into": "notifications",
      "on": [
        "user_id",
        "message"
      ]
    }
  }
])

Mongo Playground for matched / update count case
Mongo Playground for unmatched / insert case
Mongo Playground for no record / insert case