mongodb - 如何只从 Mongodb 聚合中返回一些字段（$project），同时还使用 $match、$lookup 和 $filter

问题描述

我非常接近从这个查询中得到我想要的东西......但我只想要返回一些字段，现在它正在返回所有这些字段

注意：这是一个改进：我现在询问如何仅返回某些字段，而我的类似问题询问如何返回开始日期和结束日期之间的数据

此外，有人可以请使用带有我的数据集的MongoDB Playground提供答案，以便我可以尝试一下……我不太清楚如何“命名”数据集，以便它们在操场上工作！

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)

这是一些寄存器数据

[
  {
    "_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，返回两个日期之间的数据，但是

此代码返回所有内容，包括“dontWantThis1”和“dontWantThis2”

笔记

我不希望以“dontWantThis”开头的字段 - 但这只是为了显示我不想要的字段......我并不想排除以“dontWantThis”开头的字段.....它们可能是命名为 'foo' 或 'apple' 或 'dog' 他们只是这样命名以表明我不想要它们


        // 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( ..... )

试试 1

我以为我可以做这样的事情，但它仍然返回所有通知字段

        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！！

标签： mongodbfilteringlookupaggregationprojection

mongodb - 如何只从 Mongodb 聚合中返回一些字段（$project），同时还使用 $match、$lookup 和 $filter

问题描述

注册架构

通知

有效，返回两个日期之间的数据，但是

笔记

试试 1

解决方案

解决方案

推荐阅读