Question

我正在编写大于等于CREATE_DATE的第一个查询，然后写入CREATE_DATE我得到的答案是计算持续时间写另一个查询但是如何在聚合中编写单个查询。

查询1

db.lights.aggregate({
  $match: {
    CREATE_DATE: {
      $gte: ISODate("2018-01-24T20:05:30.000Z"),
      $lt: ISODate("2018-02-24T20:05:30.000Z")
    }
  }
});

结果

{
  "_id": ObjectId("5a9a74843711955836a8b4b5"),
  "SWITCHID": "Z4-W40-SS451A/4",
  "CREATE_DATE": ISODate("2018-01-24T20:05:30Z"),
  "RECEIVEDDATE": ISODate("2018-02-24T20:05:45Z"),
  "STATUS": "LIGHTS ON"
}

查询2

db.lights.aggregate([
  {
    $addFields: {
      offduration: {
        $divide: [
          {
            $subtract: [
              "$RECEIVEDDATE",
              "$CREATE_DATE"
            ]
          },
          3600000
        ]
      }
    }
  }
]);

结果

{
  "_id": ObjectId("5a9a75af3711955836a8b4c8"),
  "SWITCHID": "Z4-W40-SS451A/5",
  "CREATE_DATE": ISODate("2018-02-24T20:05:30Z"),
  "RECEIVEDDATE": ISODate("2018-02-24T20:05:45Z"),
  "STATUS": "LIGHTS ON",
  "offduration": 0.004166666666666667
}

Answer 1

您需要像这样改进查询：

db.lights.aggregate({
  $match: {
  CREATE_DATE: {
    $gte: ISODate("2018-01-24T20:05:30.000Z"),
    $lt: ISODate("2018-02-24T20:05:30.000Z")
    }
  }
},
 {
  $addFields: {
    offduration: {
      $divide: [
        {
          $subtract: [
            "$RECEIVEDDATE",
            "$CREATE_DATE"
          ]
        },
        3600000
      ]
    }
  }
},
{
$project: {
  _id: 1,
  SWITCHID: 1,
  CREATE_DATE: 1,
  RECEIVEDDATE: 1,
  STATUS: 1,
  offduration: '$offduration',
  avgduration: { $avg: "$offduration" }      
 }
});

如何在查询

1 个答案: