MongoDB 中每年和每月的聚合 [英] Aggregation per year and month in MongoDB

问题描述

说明:我有一个数组名称作为代码，我有一个查询参数名称作为year(下拉列表 2016,2017,2018,2019,....) 和 month 下拉列表为 (ALL,01,02,03,04,05,06,07,08,09,10,11,12).

Explanation: I have an Array name as code, I have query parameters name as year(drop-down list 2016,2017,2018,2019,....) and month drop-down list as (ALL,01,02,03,04,05,06,07,08,09,10,11,12).

CASE1:如果用户选择特定年份和特定月份，则相应地获取code 数组.

CASE1: If the user selects a specific year and specific month get the code array accordingly.

CASE2:如果用户选择特定年份和月份ALL，则获取code数组完整年份.

CASE2: If the user selects the specific year and from month ALL, get the code array complete year.

{
  "code":[
    {
      "k": "2016-06-18T18",
      "v": 1
    },
    {
      "k": "2016-06-18T21",
      "v": 2
    },
    {
      "k": "2016-07-19T00",
      "v": 13
    },
    {
      "k": "2015-06-19T03",
      "v": 6
    },
    {
      "k": "2015-07-19T06",
      "v": 6
    }
  ]
}

CASE 1 Expected Output IF user-selected : From year drop-down : 2016 , month : 06

CASE 1 Expected Output IF user-selected : From year drop-down : 2016 , month : 06

{
  "code":[
    {
      "k": "2016-06-18T18",
      "v": 1
    },
    {
      "k": "2016-06-18T21",
      "v": 2
    }
  ]
}

如果用户选择的情况 2 预期输出:从年份下拉列表:2016 年，月份:所有.

CASE 2 Expected Output IF user-selected : From year drop-down : 2016 , month : ALL.

{
  "code":[
    {
      "k": "2016-06-18T18",
      "v": 1
    },
    {
      "k": "2016-06-18T21",
      "v": 2
    },
    {
      "k": "2016-07-19T00",
      "v": 13
    },
  ]
}

推荐答案

• 初始化请求参数

var year = req.year;
var month = req.month;

• 设置默认年份条件

var condition = [{
  $eq: [{ $substr: ["$$this.k", 0, 4] }, String(year)]
}];

• 检查月份是否可用，而不是全部，然后添加条件

if (month && month != "all") {
  condition.push({
    $eq: [{ $substr: ["$$this.k", 5, 2] }, String(month)] 
  });
}

• $filter 迭代代码循环并过滤以上准备好的条件
• $filter to iterate loop of code and filter above prepared conditions

db.collection.aggregate([
  {
    $set: {
      code: {
        $filter: {
          input: "$code",
          cond: { $and: condition }
        }
      }
    }
  }
])

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