mongodb 如何在mongo中计算多个字段？

x33g5p2x 于 2023-06-05 发布在 Go

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我正在尝试编写一个mongo查询，以帮助从数据生成一个报告。
文档看起来像这样：

[{
    "DeviceType" : "A",
    "DeviceStatus" : "On"
},
{
    "Device Type" : "B",
    "Device Status" : "On"
},
{
    "DeviceType" : "A",
    "DeviceStatus" : "Off"
},
{
    "DeviceType" : "A",
    "DeviceStatus" : "On"
}]

DeviceType字段可以采用任何字符串值，DeviceStatus字段可以是On、Off或InRepair中的任何一个。我想写一个mongo查询，显示每个设备的DeviceStatus计数。结果应该是这样的：

[{
    "DeviceType" : "A",
    "DeviceStatusOnCount" : "15",
    "DeviceStatusOffCount" : "13",
    "DeviceStatusInRepairCount" : "12",
},
{
    "DeviceType" : "B",
    "DeviceStatusOnCount" : "6",
    "DeviceStatusOffCount" : "14",
    "DeviceStatusInRepairCount" : "2",
}]

我怎样才能在mongo做到这一点呢？
我当前的查询只能基于DeviceType进行分组：

db.Collection_Name.aggregate([
    { $group: { _id: "$DeviceType", count: { $sum: 1 } } }
])

mongodb

来源：https://stackoverflow.com/questions/76392047/how-can-i-count-over-multiple-fields-in-mongo

1条答案

按热度按时间

bwitn5fc1#

您可以使用**$cond**+$eq向**$sum**运算符返回1或0：

db.collection.aggregate([
  {
    $group: {
      _id: "$DeviceType",
      DeviceStatusOnCount: { $sum: { $cond: [ { $eq: [ "$DeviceStatus", "On" ] }, 1, 0 ] } },
      DeviceStatusOffCount: { $sum: { $cond: [ { $eq: [ "$DeviceStatus", "Off" ] }, 1, 0 ] } },
      DeviceStatusInRepairCount: { $sum: { $cond: [ { $eq: [ "$DeviceStatus", "InRepair" ] }, 1, 0 ] } }
    }
  },
  {$addFields: {DeviceType: '$_id'}}, {$project: {_id:0}}
])

测试：mongoplayground

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mongodb 如何在mongo中计算多个字段？

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