Question

我有一个带有简单文档的Collection，用于存储impressions和conversions以及以下结构

/* 1 */
{
    "_id" : ObjectId("566f1ef857c1e6dd3123050a"),
    "path_id" : ObjectId("55944c1ebe244fd19cbf510b"),
    "data_type" : "impression",
    "created_at" : ISODate("2015-12-14T19:56:40.100Z"),
    "__v" : 0
}

/* 2 */
{
    "_id" : ObjectId("566f1fc9ac964e6f327c55d6"),
    "path_id" : ObjectId("55944c1ebe244fd19cbf510b"),
    "data_type" : "conversion",
    "created_at" : ISODate("2015-12-14T20:00:09.972Z"),
    "__v" : 0
}

/* 3 */
{
    "_id" : ObjectId("566f2896739f6afa4485f327"),
    "path_id" : ObjectId("562e594315ef3d8c3f05d219"),
    "data_type" : "impression",
    "created_at" : ISODate("2015-12-14T20:37:42.139Z"),
    "__v" : 0
}

/* 4 */
{
    "_id" : ObjectId("566f28e5739f6afa4485f328"),
    "path_id" : ObjectId("562e594315ef3d8c3f05d219"),
    "data_type" : "impression",
    "created_at" : ISODate("2015-12-14T20:39:01.233Z"),
    "__v" : 0
}

我可以按data_type进行分组和计数，但我需要做的是按date分组，然后计算data_type以获得以下结果< / p>

[
    {
        '_id': 'Y',
        'conversions': 20,
        'impressions': 2703,
        'date': '2015-12-14'
    },
    {
        '_id': 'Z',
        'conversions': 10,
        'impressions': 1703,
        'date': '2015-12-13'
    } 
]

我现在拥有的代码如下，但只有data_type的组我试图添加一个项目按日期重新组合，到目前为止没有运气。

var path_id = new mongoose.Types.ObjectId( req.body.path_id );
var match = {
    'path_id': {
        $eq: path_id
    }
};

var group = {
    '_id': '$data_type',
    'count': {
        '$sum': 1
    }
}

Hit.aggregate( [ {
    $match: match
}, {
    $group: group
} ], function( err, res ) {
    console.log( res );
} );

结果是

POST /api/hits/bypath 200 30ms - 15b
[ { _id: 'conversion', count: 2 },
  { _id: 'impression', count: 2703 } ]

任何人都可以指出我正确的道路。

先谢谢

Answer 1

要按日期嵌套分组，您必须使用日期聚合运算符 $dateToString。

这是查询

db.hits.aggregate([
    {
      "$project": {
        "created_at": {
          "$dateToString": {
            "format": "%Y-%m-%d",
            "date": "$created_at"
          }
        },
        "data_type": true
      }
    },
    {
      "$group": {
        "_id": {
          "data_type": "$data_type",
          "created_at": "$created_at"
        },
        "count": {
          "$sum": 1
        }
      }
    },
    {
      "$group": {
        "_id": {
          "data_type": "$_id.data_type"
        },
        "data":{ "$addToSet" : { count: "$count", date: "$_id.created_at" } }
      }
    }
  ])

如果您想在分组之前根据条件进行匹配，请在查询中添加以下内容

{
  "$match": {
    "path_id": {
      "$eq": "<path_id>"
    }
  }
}

Answer 2

您可以使用Date Aggregation Operators投影日/月/年字段，然后按其分组

{
  "$project": {
    "y": {
      "$year": "$created_at"
    },
    "m": {
      "$month": "$created_at"
    },
    "d": {
      "$dayOfMonth": "$created_at"
    },
    "data_type" : 1
  }
},
{
  "$group": {
    "_id": {
      "year": "$y",
      "month": "$m",
      "day": "$d",
      "data_type": "$data_type"
    },
    count: {
      "$sum": 1
    }
  }
}

并将以此格式输出：

    "_id": {
      "year": 2015,
      "month": 10,
      "day": 5,
      "data_type": "impression"
    },
    count: 10

然后按日期再次分组以合并一个文档中的类型

{
  "$group": {
    "_id": {
      "year": "$_id.year",
      "month": "$_id.month",
      "day": "$_id.day"
    },
    types: {"$push":"$_id.data_type"},
    counters: {"$push":"$count"}
  }
}

将导致此：

    "_id": {
      "year": 2015,
      "month": 10,
      "day": 5
    },
    types: ["impression", "conversion"]
    counters: [10, 5]

虽然可能有更优雅或更快（有1组）的方式，但我不确定。

MongoDB按类型和按日期分组

2 个答案: