mongo 聚合添加日期和总和

Question

我觉得这可能真的很复杂。我知道这在 JS 中是可能的，方法是获取每条记录，确定月份差异，然后在 for 循环中将每个 model.month 增加 1，一旦满足月份差异，循环就会停止。我只是无法理解聚合的可能性。任何帮助都是救命稻草！

这是我的数据：

{
   model: car
   serial: bbbaaa
   warranty_start_date: 03/05/2018
   warranty_end_date: 06/16/2018
},
{
   model: car
   serial: jjjwww
   warranty_start_date: 03/02/2018
   warranty_end_date: 05/20/2018
},
{
   model: truck
   serial: tttvvv
   warranty_start_date: 05/06/2016
   warranty_end_date: 07/15/2016
}

这就是我想要的结局：

{
  model: car
  in_warranty_these_months: { 03/2018: 2, 04/2018: 2, 05/2018: 2, 06/2018: 1 } 
},
{
  model: truck
  in_warranty_these_months: { 05/2016: 1, 06/2016: 1, 07/2016: 1 } 
}

更新在@mickl 的大力帮助下，下面的代码就像一个魅力：

 db.col.aggregate([
    {
        $addFields: {
            monthsRange: {
                $range: [ 
                    { $add: [ 
                        { $multiply: [12, {$year: "$warranty_start_date"}] }, 
                        {$month: "$warranty_start_date"} ] 
                    }, 
                    { $add: [ 
                        { $multiply: [12, {$year: "$warranty_end_date"}] }, 
                        { $add: [{$month: "$warranty_end_date"}, 1] } ] 
                    }, 1]
            }
        }
    },
    {
        $unwind: "$monthsRange"
    },
    {
        $group: {
            _id: { model: "$model", month: "$monthsRange" },
            count: {$sum:1}
        }
    },
    {
      $group: {
        _id: "$_id.model",
        pairs: {
          $push: {          
            k: {
              $dateToString: {
                format: "%m_%Y",
                date: {
                  $dateFromParts: {
                    day: 1,
                    month: {
                      $cond: [{
                        $eq: [0, {
                          $mod: ["$_id.month", 12]
                        }]
                      }, {
                        $trunc: 12
                      }, {
                        $mod: ["$_id.month", 12]
                      }]
                    },
                    year: {
                      $cond: [{
                        $eq: [0, {
                          $mod: ["$_id.month", 12]
                        }]
                      }, {
                        $subtract: [{
                          $trunc: {
                            $divide: ["$_id.month", 12]
                          }
                        }, 1]
                      }, {
                        $trunc: {
                          $divide: ["$_id.month", 12]
                        }
                      }]
                    }
                  }
                }
              }
            },
            v: "$count"
          }
        }
      }
    },
    {
            $project: {
                _id: 0,
                model: "$_id",
                in_warranty_these_months: {
                    $arrayToObject: "$pairs"
                }
            }
        }
    ])

Answer 1

您可以使用以下聚合：

db.col.aggregate([
    {
        $addFields: {
            monthsRange: {
                $range: [ 
                    { $add: [ 
                        { $multiply: [12, {$year: "$warranty_start_date"}] }, 
                        {$month: "$warranty_start_date"} ] 
                    }, 
                    { $add: [ 
                        { $multiply: [12, {$year: "$warranty_end_date"}] }, 
                        { $add: [{$month: "$warranty_end_date"}, 1] } ] 
                    }, 1]
            }
        }
    },
    {
        $unwind: "$monthsRange"
    },
    {
        $group: {
            _id: { model: "$model", month: "$monthsRange" },
            count: {$sum:1}
        }
    },
    {
        $group: {
            _id: "$_id.model",
            pairs: {
                $push: {
                    k: { 
                        $dateToString: {
                            format: "%m/%Y",
                            date: {
                                $dateFromParts: { 
                                    day: 1, 
                                    month: { $add: [{ $mod: [ "$_id.month", 12 ] }, 1] }, 
                                    year: { $trunc: { $divide: [ "$_id.month", 12 ] } }
                                } 
                            }
                        }                        
                    },
                    v: "$count"
                }
            }
        }
    },
    {
        $project: {
            _id: 0,
            model: "$_id",
            in_warranty_these_months: {
                $arrayToObject: "$pairs"
            }
        }
    }
])

基本上，您必须根据日期生成范围（使用 $range）。为此，您可以根据以下公式将日期转换为数字：12 *year + month。这将使您有可能使用 $range 为第一个文档生成四个值，为第二个文档生成三个值，等等。

然后你可以使用$group来计算每个模型的每个月。

最后一步我们要使用$arrayToObject so we have to transform our data to objects with two properties, k and v. To do that we have to transform our numbers back to desired format using $dateFromParts and $dateToString

mongo 聚合添加日期和总和

mongo aggregation adding dates and sums

javascript

aggregation

mongoose

mongodb

node.js