Mongodb mongo多重计数和设备结果
我有一个mongodb系列,如下所示:Mongodb mongo多重计数和设备结果,mongodb,mongodb-query,aggregation-framework,aggregate,Mongodb,Mongodb Query,Aggregation Framework,Aggregate,我有一个mongodb系列,如下所示: [ { "_id" : ObjectId("5ba0e5a99e7537012371855a"), "user_id" : 2, "action" : 0, "source" : 1, "service" : "FootPlus", "free" : false, "source_detail" : { "source_type" : "landi
[
{
"_id" : ObjectId("5ba0e5a99e7537012371855a"),
"user_id" : 2,
"action" : 0,
"source" : 1,
"service" : "FootPlus",
"free" : false,
"source_detail" : {
"source_type" : "landing",
"source_id" : 2,
"promoter_id" : 1
},
"created_at" : ISODate("2018-09-18T11:46:49.000Z")
}
{
"_id" : ObjectId("5ba0e5cc9e7537013d57e37a"),
"user_id" : 2,
"action" : 1,
"service" : "FootPlus",
"source" : 0,
"created_at" : ISODate("2018-09-18T11:47:24.000Z"),
"source_detail" : {
"source_type" : "landing",
"source_id" : 2,
"promoter_id" : 1
}
}
]
我想按source\u detail.promotor\u id和count action=0和action=1进行分组
并将每个启动子的作用计数=0除以作用计数=1
因此,我的结果应该是:
[
{
"promoter_id": 2
"action_0": 27,
"action_1": 9,
"devide": 3
},
{
"promoter_id": 3
"action_0": 18,
"action_1": 3,
"devide": 6
}
]
我怎样才能达到这个结果?
谢谢。您可以使用下面的聚合 差不多
db.colname.aggregate([
{"$group":{
"_id":"$source_detail.promoter_id",
"action_0":{"$sum":{"$cond":[{"$eq":["$action",0]},1,0]}},
"action_1":{"$sum":{"$cond":[{"$eq":["$action",1]},1,0]}}
}},
{"$addFields":{"divide":{"$divide":["$action_0","$action_1"]}}}
])