How to get the total number of times each object has been referenced in a collection? - mongodb

Question

So what i am trying to find is how many time each kite was flown by each person, the db looks like this:
{
"_id" : ObjectId("5bccf4d7ac8fa95f43a964c2"),
"KITE" : {
"registration" : "DTH498"
}
}
{
"_id" : ObjectId("5bccf4d7ac8fa95f43a964c3"),
"KITE" : {
"registration" : "HKJ607"
}
}
{
"_id" : ObjectId("5bccf4d7ac8fa95f43a964c4"),
"KITE" : {
"registration" : "GCF21"
}
}
{
"_id" : ObjectId("5bccf4d7ac8fa95f43a964cf"),
"PERSON" : {
"name" : "H.Y",
"used" : [
{
"registration" : "DTH498"
},
{
"registration" : "HKJ607"
},
{
"registration" : "GCF21"
}
]
}
}
{
"_id" : ObjectId("5bccf4d7ac8fa95f43a9leo5"),
"PERSON" : {
"name" : "T.G",
"used" : [
{
"registration" : "DTH498"
},
{
"registration" : "HKJ607"
},
{
"registration" : "GCF21"
}
]
}
}
{
"_id" : ObjectId("5bccf4d7ac8fro4943a01pak"),
"PERSON" : {
"name" : "X.L",
"used" : [
{
"registration" : "DTH498"
},
{
"registration" : "HKJ607"
},
{
"registration" : "GCF21"
}
]
}
}
Right now i can only use the aggregation framework, with that said i've managed to list all kites that have been used:
db.data.aggregate([
{"$unwind":"$PERSON.used"},
{"$group":{"_id":"$PERSON.used.registration"}}
]).pretty();
The result i'm trying to get to would have the registration of the kite and the number of times it has been used overall.
I'm not sure how it would look but, what im thinking is:
1. Somehow get all used array objects registrations into a new array so it one big list of all the times each kite has been used
2. Group the array by registration and use sum to display how many duplicates each group had.

Answer 1

I DID IT, WOOOO!
so if anyone ever stumbles upon my question, heres the answer.
First unwind the used array
Second project only the registrations
Third group with an _id of registration and count with $sum
db.data.aggregate([
{"$unwind":"$PERSON.used"},
{"$project":{"registration":"$PERSON.used.registration","_id":0}},
{"$group":{"_id":"$registration", "count":{"$sum":1}}}
]);