我如何只使用 Mongodb 聚合中的 return 某些字段（$project），同时还使用 $match、$lookup 和 $filter

Question

我非常接近从这个查询中得到我想要的东西......但我只想要一些字段 returned 而现在它正在 returning 所有这些

注意：这是一个改进：我现在问如何return 只有某些字段，而我的类似询问如何 return 数据 在开始日期和结束日期之间

此外，有人可以使用 MongoDB Playground 和我的数据集提供答案吗？这样我就可以试试了...我不太明白如何 "name"数据集让他们在操场上工作！

注册架构

const mongoose = require('mongoose');
const Schema = mongoose.Schema;
const RegisterSchema = new Schema({
    userId: {type: Schema.Types.ObjectId, required: true},
    accessToken: {type:String, required: true, default: null},
})
module.exports = Register = mongoose.model( 'register', RegisterSchema)

这是一些注册数据

[
  {
    "_id": "5eac9e815fc57b07f5d0d29f",
    "userId": "5ea108babb65b800172b11be",
    "accessToken": "111"
  },
  {
    "_id": "5ecaeba3c7b910d3276df839",
    "userId": "5e6c2dddad72870c84f8476b",
    "accessToken": "222"
  }
]

下一个文档包含通过 accessToken 与 Register 模式相关的数据

通知

const mongoose = require('mongoose');
const Schema = mongoose.Schema;
const NotificationSchema = new Schema({
    accessToken: {type:String, required: true},
    summaryId: {type:Number, required: true},
    dateCreated: {type: Date, default: Date.now},

    // I don't want these returned in the final results
    dontWantThis1: {type:Number, required: true},
    dontWantThis2: {type:Number, required: true},
})
module.exports = Notification = mongoose.model( 'notification', NotificationSchema)

这是一些通知数据

[{
    "_id": "5ebf0390c719e60004f42e74",
    "accessToken": "111",
    "summaryId": 1111,
    "dontWantThis1": 61,
    "dontWantThis2": 62,
    "dateCreated": "2020-04-17T00:00:00.000+00:00" }, 
  {
    "_id": "6ebf0390c719e60004f42e76",
    "accessToken": "222",
    "summaryId": 2221,
    "dontWantThis1": 71,
    "dontWantThis2": 72,
    "dateCreated": "2020-04-18T00:00:00.000+00:00" },
  {
    "_id": "6ebf0390c719e60004f42e78",
    "accessToken": "111",
    "summaryId": 1112,
    "dontWantThis1": 611,
    "dontWantThis2": 622,
    "dateCreated": "2020-05-25T00:00:00.000+00:00" },
  {
    "_id": "6ebf0390c719e60004f42e80",
    "accessToken": "222",
    "summaryId": 2222,
    "dontWantThis1": 711,
    "dontWantThis2": 722,
    "dateCreated": "2020-05-26T00:00:00.000+00:00" }
]

有效，两个日期之间的 returns 数据，但是

此代码 return 包含所有内容，包括 'dontWantThis1' 和 'dontWantThis2'

注意

我不想要以 'dontWantThis' 开头的字段 - 但这只是为了显示我不想要的字段......我不是字面意思想要排除以 'dontWantThis' 开头的字段......它们可以命名为 'foo' 或 'apple' 或 'dog' 它们只是以这种方式命名以表明我不想要他们


        // make sure the input dates are REALLY date objects
        // I only want to see notifications for the month of May (in this example)
        var dateStart = new Date('2020-05-01T00:00:00.000+00:00');
        var dateEnd = new Date('2020-05-30T00:00:00.000+00:00');     

        var match = {$match: { userId: mongoose.Types.ObjectId(userId) } };

        var lookup ={
            $lookup:
            {
                from: "my_Notifications",
                localField: "accessToken",
                foreignField: "accessToken",
                as: "notifications"
            }
        };

        var dateCondition = { $and: [
            { $gte: [ "$$item.dateCreated", dateStart ] },
            { $lte: [ "$$item.dateCreated", dateEnd ] }
          ]}  

        var project = {
            $project: {
                notifications: {
                    $filter: {
                    input: "$notifications",
                    as: "item",
                    cond: dateCondition
                    } } }
        };

        var agg = [
            match,
            lookup,
            project
        ];

        Register.aggregate(agg)
        .then( ..... )

尝试 1

我以为我可以做这样的事情，但它仍然 return所有通知字段

        var project = {
            $project: {
                "_id": 1,
                "userId": 1,
                "accessToken":1,
                "count":{$size:"$notifications"},
                "notifications._id":1,
                "notifications.summaryId": 1,
                "notifications.dateCreated":1,

                notifications : {
                    $filter: {
                    input: "$notifications",
                    as: "item",
                    cond: dateCondition
                    },
            }}
        };

解决方案

我创建了另一个投影并将其添加到管道中：

        var project2 = {
            $project: {
                "_id": 1,
                "userId": 1,
                "accessToken":1,
                "count":{$size:"$notifications"},
                "notifications._id":1,
                "notifications.summaryId": 1,
                "notifications.dateCreated":1,
                "notifications.dateProcessed":1,
            }
        };


        var agg = [
            match,
            lookup,
            project,
            project2,
        ];

谢谢！！

Answer 1

https://whosebug.com/users/6635464/ngshravil-py 准确无误。

我创建了另一个投影：

        var project2 = {
            $project: {
                "_id": 1,
                "userId": 1,
                "accessToken":1,
                "count":{$size:"$notifications"},
                "notifications._id":1,
                "notifications.summaryId": 1,
                "notifications.dateCreated":1,
                "notifications.dateProcessed":1,
            }
        };

然后将其添加到我的聚合管道中：

        var agg = [
            match,
            lookup,
            project,
            project2,
        ];

成功了！ -- 谢谢 https://whosebug.com/users/6635464/ngshravil-py

Answer 2

您需要使用 $dateFromString 和 $map 运算符将 notifications.dateCreated 转换为 ISODate，因为您的日期是字符串形式。我建议你这样做，因为我不认为你可以用字符串格式进行日期比较。另外，请确保 dateStart 和 dateEnd 也应采用 ISODate 格式。

并且您需要两个 $project 运算符才能实现此目的。此外，我没有看到任何带有 userAccessToken 的字段，我假设它是 accessToken。检查以下查询。

db.Register.aggregate([
  {
    $lookup: {
      from: "my_Notifications",
      localField: "accessToken",
      foreignField: "accessToken",
      as: "notifications"
    }
  },
  {
    $project: {
      "_id": 1,
      "userId": 1,
      "accessToken": 1,
      notifications: {
        $map: {
          input: "$notifications",
          as: "n",
          in: {
            "_id": "$$n._id",
            "summaryId": "$$n.summaryId",
            "dateCreated": {
              $dateFromString: {
                dateString: "$$n.dateCreated"
              }
            }
          }
        }
      }
    }
  },
  {
    $project: {
      "userId": 1,
      "accessToken": 1,
      "notifications": {
        $filter: {
          input: "$notifications",
          as: "item",
          cond: {
            $and: [
              {
                $gte: [
                  "$$item.dateCreated",
                  ISODate("2020-05-24T00:00:00Z")
                ]
              },
              {
                $lte: [
                  "$$item.dateCreated",
                  ISODate("2020-05-26T00:00:00Z")
                ]
              }
            ]
          }
        }
      }
    }
  },
  {
    $set: {
      "count": {
        $size: "$notifications"
      }
    }
  }
])

MongoPlayGroundLink

我如何只使用 Mongodb 聚合中的 return 某些字段（$project），同时还使用 $match、$lookup 和 $filter

How do I only return SOME fields ($project) from Mongodb aggregation, while also using $match, $lookup AND $filter

mongodb

filtering

lookup

aggregation

projection

注册架构

通知

有效，两个日期之间的 returns 数据，但是

注意

尝试 1

解决方案