在mongodb聚合中计算另一个文档中的项目 [英] Count items from another document in mongodb aggregation

问题描述

我有两个不同的文档

Accounts
[
  {_id:1, name:'xyz'},
  {_id:2, name:'abc'},
  {_id:3, name:'def'},
  {_id:4, name:'pqr'},
]

和

Responses
[
  {_id:01, accountId:2, questionId: 0001, res: true},
  {_id:02, accountId:2, questionId: 0002, res: true},
  {_id:03, accountId:1, questionId: 0003, res: false},
  {_id:03, accountId:3, questionId: 0002, res: false},
  {_id:03, accountId:2, questionId: 0003, res: false},
]

我如何计算单个帐户的 true 和 false 响应数，同时也要保留其原始字段.

How can I count number of true and false responses for an individual account while maintaining its original fields too.

例如.

{
  _id: 2,
  name: 'abc',
  trueResponses: 2,
  falseResponses: 1
}

我曾尝试使用$ lookup在字段中加入其他文档，但无法计算不同的响应.

I have tried using $lookup to join the other document in a field but am unable to count different responses.

db.accounts.aggregate([
    {
        $match: { _id: 2 }
    },
    {
        $lookup:{
            from: 'responses',
            localField: '_id',
            foreignField: 'accountId',
            as: 'responses'
        }
    }
])

推荐答案

有很多方法，

• $ match 消除不需要的数据
• $ lookup 加入收藏集
• $ unwind 解构数组
• $ group 重建数组，而 $ sum 使用 $ cond 条件
帮助增加1

• $match eliminate unwanted data
• $lookup to join collections
• $unwind to deconstruct the array
• $group to reconstruct the array and $sum helps to increase by 1 using $cond condition

这是脚本

db.Accounts.aggregate([
  { $match: { _id: 2 } },
  {
    $lookup: {
      from: "Responses",
      localField: "_id",
      foreignField: "accountId",
      as: "responses"
    }
  },
  {
    $unwind: "$responses"
  },
  {
    $group: {
      _id: "$_id",
      name: { $first: "$name" },
      trueResponses: {
        $sum: {
          $cond: [{ $eq: [ "$responses.res", true]},1,0]
        }
      },
      falseResponses: {
        $sum: {
          $cond: [{ $eq: [ "$responses.res", false]},1,0]
        }
      }
    }
  }
])

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