如何使用聚合过滤 mongodb 两个日期之间返回的数据：匹配、查找和项目？

Question

我想在传入 userId 时获取开始日期和结束日期之间与注册相关的所有通知。

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},
})
module.exports = Notification = mongoose.model( 'notification', NotificationSchema)

这是一些通知数据

[{
    "_id": "5ebf0390c719e60004f42e74",
    "accessToken": "111",
    "summaryId": 1111,
    "dateCreated": "17 Apr 2020" }, 
  {
    "_id": "6ebf0390c719e60004f42e76",
    "accessToken": "222",
    "summaryId": 2221,
    "dateCreated": "18 Apr 2020" },
  {
    "_id": "6ebf0390c719e60004f42e78",
    "accessToken": "111",
    "summaryId": 1112,
    "dateCreated": "25 May 2020" },
  {
    "_id": "6ebf0390c719e60004f42e80",
    "accessToken": "222",
    "summaryId": 2222,
    "dateCreated": "26 May 2020" }
]

尝试 1

        var userId = '5ea108babb65b800172b11be'
        var dateStart = '27 Apr 2020';
        var dateEnd   = '27 May 2020'; 

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

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

        project = {
            $project: {
                items: {
                    $filter: {
                    input: "$items",
                    as: "item",
                    cond: { {"dateCreated": {'$gte': dateStart, '$lte': dateEnd }} }
                    }
                }
            }
        };

        var agg = [
            match,
            lookup,
            project
        ];


        Register.aggregate(agg)
        .then( events => {
            if(events){
                return resolve(events);
            }else{
                return reject({success:false});
            }
        })
        .catch(err => {
            console.log('ERROR ' + JSON.stringify(err.message));
            return reject({success:false});
        })

尝试 1 次错误

我希望看到 5 月 25 日的 accessToken 为 111 的通知，但出现错误：

ERROR : {"\"An object representing an expression must have exactly one field: { $gte: new Date(1588017802546), $lte: new Date(1590609802546) }\""}

尝试 2

我摆脱了错误......但仍然没有得到任何回报：

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

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

这是我的项目的样子：

{
  "$project": {
    "items": {
      "$filter": {
        "input": "$items",
        "as": "item",
        "cond": {
          "$and": [
            {"$gte": ["$$item.dateCreated",1588019227296] },
            {"$lte": ["$$item.dateCreated",1590611227296] }
          ] } } } }
}

尝试 3

使用评论中的建议...我将 'items'（从尝试 2）更改为 'notifications'

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

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

还是不行

因此，为了尽可能简化它以使其正常工作...我正在尝试使用摘要 ID

尝试 4

          dateCondition = { $and: [
            { $gte: [ "$$item.summaryId", 1 ] },
            { $lte: [ "$$item.summaryId", 555555 ] }
          ] }          

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

有效...所以我认为这是日期问题。

最终代码 - 有效！


        // make sure the input dates are REALLY date objects
        var dateStart = new Date(inputDateStart);
        var dateEnd = new Date(inputDateEnd);     

        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 ] }
          ]}  

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

        var agg = [
            match,
            lookup,
            project
        ];

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

Answer 1

如果 dateStart 和 dateStart 实际上是 Date 个对象而不是 String 个对象，您的解决方案看起来几乎是正确的。

您的 Try 2 不完整我不确定它是否使用 Try 1 中的 $lookup。如果是这样，您必须确保 $lookup 的输出与 $filter 的输入相同。因此，您应该更改 $lookup 中的 as 以匹配 $filter

的 input

{
  $lookup: {
    from: "notifications",
    localField: "accessToken",
    foreignField: "accessToken",
    as: "items" // here
  }

}

备选方案

我不确定你想要什么作为输出。如果您只需要通知数组而不需要用户对象，您可以尝试以下方法。

[{
  $match: { userId: mongoose.Types.ObjectId(userId) }
}, {
  $lookup: {
    from: "notifications",
    localField: "accessToken", // don't forget to index register.accessToken
    foreignField: "accessToken", // don't forget to index notification.accessToken
    as: "notifications"
  }
}, {
  $unwind: "$notifications"
}, {
  $match: { 
    dateCreated: { $gte: dateStart, $lte: dateEnd } // dateStart, dateEnd should be Date objects
  }
}, { // optional, move notifications to top lvel
  $replaceRoot: { root: '$notifications' }
}]

如何使用聚合过滤 mongodb 两个日期之间返回的数据：匹配、查找和项目？

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