简体   繁体   English

Mongo Aggregate - 根据日期计算

[英]Mongo Aggregate - Calculating based on dates

Here is a sample document.这是一个示例文档。

{"_id":{"$oid":"5e557779ed588826d84cef11"},
"meter_id":"1001",
"date":{"$date":{"$numberLong":"1509474600000"}},
"parameter_name":"hvac","voltage":{"unit":"V"},
"current":{"unit":"AMP"},
"powerFactor":{"unit":"phi"},
"angle":{"unit":"degree"},
"activePower":{"unit":"kwh"},
"reactivePower":{"unit":"kwh"},
"apparentPower":{"unit":"kwh"},
"frequency":{"unit":"hz"},
"thd":{"unit":"percentage"},
"energy":{"Energy":"5.7"},
"power":{"unit":"watt"},

And there are around 100 000 documents.并且有大约 100 000 个文件。 I want to filter out the documents whose date is greater than the date i specify and calculate the total energy ie energy.Energy我想过滤掉日期大于我指定日期的文件并计算总能量,即energy.Energy

I used the below aggregation, but it doesn't seem to be working我使用了下面的聚合,但它似乎不起作用

const endDate = new Date(12-12-2018)
 MeterData.aggregate([
      {
          $group: {
              _id: "$meter_id",
              total: { $sum: 1 },
              totalEnergy: { $sum: { $toDouble: "$energy.Energy"
                   } },
                   dateSum: {
                    $sum: {
                      $toDouble: {
                        $not: [{
                          $cond: {
                            if: {
                              $gte: [
                                "$date", endDate
                              ]
                            },
                            then: "$energy.Energy",
                            else: 0
                          }
                        }

                        ]


                    }
                    }
                   }
          }
      }
  ])

Would be this:会是这样:

db.collection.aggregate([
   { $match: { date: { $gte: ISODate("2016-12-12") } } },
   {
      $group: {
         _id: "$meter_id",
         total: { $sum: 1 },
         totalEnergy: { $sum: "$energy.Energy" }
      }
   }
])

See Mongo playground参观Mongo 游乐场

声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM