mongodb - 查询有n条记录的集合

Question

• 我有一个名为 GPS 的集合，每秒有 n 个与服务器连接的设备将数据推送到集合。样本数据

 {
     "_id" : ObjectId("5dd2d4f7d092fb70c022c700"),
     "date" : "2019-11-18",
     "altitude" : "0.0",
     "latdegree" : "N",
     "gpspositioning" : "V",
     "positionType" : "GPS",
     "latitude" : "12.900660",
     "range" : null,
     "gsmsignalstrength" : "46",
     "battery" : "87",
     "terminalstatus" : "00000000",
     "longitude" : "80.2347400",
     "satellites" : "0",
     "created" : "2019-11-18 17:29:26.991Z",
     "longdegree" : "E",
     "tumblingtimes" : "0",
     "positionaccuracy" : "52",
     "time" : "17:30:01.000Z",
     "locationDate" : "2019-11-18 17:30:01.000Z",
     "device" : "6005868660"
 }

所以现在我想用每个设备的最后更新值查询这个集合，即设备的数量作为数组。例如

[
  {
    device:1,
    lat:11,
    long:4111
  },
  {
    device:2,
    lat:11,
    long:4111
  },
  {
    device:3,
    lat:11,
    long:4111
  }
.....n]

我尝试过使用 find 和 $natural 它查找整个数据而不是最后更新的数据。请问我该怎么做？谢谢

Answer 1

我猜你需要aggregation ，尝试类似的事情。我在这里所做的是首先使用 locationDate 按升序对数据进行排序( $sort :1)(我猜它是设备的最后一个位置时间戳)，然后在现场应用分组device 然后选择管道中的最后一个数据。您还可以执行 $sort:-1 (降序排序并选择 $first)

$last 是什么意思？做什么？

Returns the value that results from applying an expression to the last document in a group of documents that share the same group by a field. Only meaningful when documents are in a defined order.

 db.gps.aggregate(
       [
         { $sort: { locationDate: -1 } },
         {
           $group:
             {
               _id: "$device",
               latitude:{$first: "$latitude"},
               longitude: {$first: "$longitude"},
               lastupdateValue: { $first: "$locationDate" }
             }
         }
       ]
    )

或者

     db.gps.aggregate(
           [
             { $sort: { locationDate: 1 } },
             {
               $group:
                 {
                   _id: "$device",
                   latitude:{$last: "$latitude"},
                   longitude: {$last: "$longitude"},
                   lastupdateValue: { $last: "$locationDate" }
                 }
             }
           ]
        )