Mongo 查询聚合条件组和另一个字段

Question

我关注了@Blakes-Seven 的这篇文章 Mongodb 聚合多个日期范围的$group

但是我需要按字段“$User.Account”添加一个额外的组，并且我不断收到错误。当我把它拿出来时，效果很好。

我正在尝试做的事情是找到每个日期范围内的前 N 个用户......

{
    "message" : "the group aggregate field 'User' must be defined as an expression inside an object",
    "ok" : 0,
    "code" : 15951,
    "name" : "MongoError"
}

任何帮助将不胜感激。我错过了一些东西...

>     // work out dates somehow var today = new Date(),
>     oneDay = ( 1000 * 60 * 60 * 24 ),
>     ninetyDays = new Date( today.valueOf() - ( 90 * oneDay ) ),
>     sixtyDays = new Date( today.valueOf() - ( 60 * oneDay ) ),
>     thirtyDays = new Date( today.valueOf() - ( 30 * oneDay ) );
> 
> db.logs.aggregate([
>     { "$match": {
>         "DateTime": { "$gte": ninetyDays },
>         "Orgname": /Inc/
>     }},
>     { "$group": {
>         "_id": {
>             "$cond": [
>                 { "$lt": [ "$DateTime", sixtyDays ] },
>                 "61-90",
>                 { "$cond": [
>                     { "$lt": [ "$DateTime", thirtyDays ] },
>                     "31-60",
>                     "01-30"
>                 ]}
>             ]
>         },
>         "User": "$User.Account",
>         "count": { "$sum": 1 },
>     }},
>     {    $sort: {"count": -1}
>     },
>     { $limit: 25} ])

Sample output
01-30  usera 45
01-30  userc 34
01-30  userf 28
01-30  userq 13
… 20 more rows...
01-30  usery 4
31-60  userb 55
… 23 more rows
31-60  userk 3
61-90  userm 78
61-90  userf 45
... 22 more rows...

61-90  usery 22

Answer 1

您可以按照以下语法在

$group

表达式中添加另一个字段

db.logs.aggregate([
  { "$match": {
    "DateTime": { "$gte": ninetyDays },
    "Orgname": /Inc/
  }},
  { "$group": {
    "_id": {
      "User": "$User.Account",
      "date": {
        "$cond": [
          { "$lt": [ "$DateTime", sixtyDays ] },
          "61-90",
          { "$cond": [
            { "$lt": [ "$DateTime", thirtyDays ] },
            "31-60",
            "01-30"
          ]}
        ]
      }
    },
    "count": { "$sum": 1 },
  }},
  { "$sort": { "count": -1 }},
  { "$limit": 25}
])

Answer 2

希望这个例子有帮助

db.collection.aggregate([
  {
    $group: {
      _id: {
        "type": "$type",
        $cond: {
          if: { $in: ["$category", [1, 2]] },
          then: "$category",
          else: "$otherField" // Replace "otherField" with the actual name of your second field## Heading ##
        }
      },
      count: { $sum: 1 }
    }
  }
])

$in 可以替换为 $eq $nin 和其他所需的约束