如何最大程度地减少此MongoDB查询？

Question

我整天都试图将所有这些查询都放在一个中，但是什么都没有用，请帮助 我需要更新计数，如果arset_id存在数组中，或插入{asset_id：id，count：1}如果不是 这是数组

{
 from: '1', 
 to: '2', 
 json:[
        {asset_id: id, count: 1},
        {asset_id: id, count: 1},
        {asset_id: id, count: 1}
 ]
}

，这是查询

      {
        updateOne: {
          filter: {
            from: res[i].from,
            to: res[i].to,
          },
          update: {
            from: res[i].from,
            to: res[i].to,
          },
          upsert: true,
        },
      },
      {
        updateOne: {
          filter: {
            from: res[i].from,
            to: res[i].to,
            json: { $elemMatch: { asset_id: res[i].asset_id } },
          },
          update: {
            $inc: { "json.$.count": 1 },
          },
        },
      },
      {
        updateOne: {
          filter: {
            from: res[i].from,
            to: res[i].to,
            json: { $not: { $elemMatch: { asset_id: res[i].asset_id } } },
          },
          update: {
            $push: {
              json: { asset_id: res[i].asset_id, count: 1 },
            },
          },
          //upsert: true,
        },
      }

Answer 1

如果阵列的顺序不重要：您可以做：

db.collection.aggregate([
  {$set: {
      jsonWithOut: {
        $filter: {
          input: "$json", cond: {$ne: ["$this.asset_id", dataToInsert.asset_id]}}
      },
      jsonOfNew: {
        $filter: {
          input: "$json", cond: {$eq: ["$this.asset_id", dataToInsert.asset_id]}}
      },
      json: "$REMOVE"
    }
  },
  {$set: {
      jsonOfNew: {
        $cond: [
          {$eq: [{$size: "$jsonOfNew"}, 0]},
          [dataToInsert],
          [
            {
              asset_id: "$dataToInsert.asset_id",
              count: {$add: [{$first: "$jsonOfNew.count"}, 1]}
            }
          ]
        ]
      },
      dataToInsert: "$REMOVE"
    }
  },
  {$set: {
      json: {$concatArrays: ["$jsonWithOut", "$jsonOfNew"]},
      jsonOfNew: "$REMOVE",
      jsonWithOut: "$REMOVE"
    }
  }
])

查看其在