MongoDB $group aggregation - mongodb

Question

I have collection like this
OrgName EmpId Domain Date
Google 12345 ABC 2017/01/01
Google 12345 XYZ 2017/02/01
Google 67890 ABC 2017/03/01
Google 45678 ABC 2017/03/02
Yahoo 69875 HGF 2017/03/02
Google 45678 XYZ 2017/03/03
Google 45678 XYZ 2017/03/03
Google 12345 XYZ 2017/03/03
Google 12345 ABC 2017/03/04
Google 12345 ABC 2017/04/05
I need to fetch which employee having the max "Domain" count and must be in both "ABC" and "XYZ" domain GROUPBY OrgName wise.
I am using below query:
db.Collection1.aggregate([{ "$match" : { "$or" : [ { "Domain": "ABC"},{ "Domain": "XYZ"}]}},
{
$group :{ "_id": {"OrgName" : "$OrgName", "EmpId" : "$EmpId",
"Domain" : "$Domain"},
            count:{ $sum : 1 },
"participantData" : { "$push" : { "EmpId" : "$EmpId" , "Domain" : "$Domain"}}}},
    {$sort:{"count":-1}},
     {$limit: 10}
],{ allowDiskUse: true })
In above example, am expecting result : employee_id=12345 present in both "ABC" and "XYZ" Domain count is 5 (i.e., 12345.ABC = 3 and 12345.XYZ=2).

Answer 1

You can try below query.
The below query $group by OrgName, EmpId followed by $match to filter documents where participant array contains both 'ABC' & 'XYZ` value.
$sort the filtered data by count and output first 10 values.
db.collection.aggregate([
{"$match":{"$or":[{"Domain":"ABC"},{"Domain":"XYZ"}]}},
{"$group":{
"_id":{"OrgName":"$OrgName","EmpId":"$EmpId"},
"count":{"$sum":1},
"participantData":{"$push":{"EmpId":"$EmpId","Domain":"$Domain"}}
}},
{"$match":{"participantData.Domain":{"$all":["ABC","XYZ"]}}},
{"$sort":{"count":-1}},
{"$limit":10}
])