How to join two Aggregation results in MongoDB? - mongodb

Question

I have a data set looks as
{"BrandId":"a","SessionId":100,"Method": "POST"}
{"BrandId":"a","SessionId":200,"Method": "PUT"}
{"BrandId":"a","SessionId":200,"Method": "GET"}
{"BrandId":"b","SessionId":300,"Method": "GET"}
I wrote aggregation count distinct session id by brandid:
db.collection.aggregate([
{$group: {
"_id": {
brand: "$BrandId",
session: "$SessionId"
},
count: {$sum: 1}
}},
{$group: {
_id: "$_id.brand",
countSession:{$sum:1}
}}
])
The expected result of the query is :
{ "_id" : "a", "countSession" : 2 }
{ "_id" : "b", "countSession" : 1 }
Another query is to count where the Method is POST by brand:
db.collection.aggregate([
{$match: {Method:"POST"}},
{$group: {
_id: '$BrandId',
countPOST:{$sum:1}
}}
])
The expected result:
{ "_id" : "a", "countPOST" : 1 }
{ "_id" : "b", "countSession" : 0 }
And now, I want to combine these two query and get the expected result as following:
{"BrandId:"a","countSession":2,"countPOST":1}
{"BrandId:"b","countSession":1,"countPOST":0}
I do not how to combine these two result of two aggregation, anyone can help?

Answer 1

You can use $cond operator as follows.
db.Collection.aggregate(
{
'$group': {
'_id': {'BrandId':'$BrandId','Session': '$SessionId'},
'countPOST':{
'$sum':{
'$cond': [{'$eq':['$Method','POST']},1,0]
}
}
}
},
{
'$group': {
'_id': '$_id.BrandId',
'countSession': {'$sum':1},
'countPOST': {'$sum': '$countPOST'}
}
}
)
Ouput:
{
"result" : [
{
"_id" : "a",
"countSession" : 2,
"countPOST" : 1
},
{
"_id" : "b",
"countSession" : 1,
"countPOST" : 0
}
],
"ok" : 1
}