SQL to Mongo Aggregation - mongodb

Question

Hi I want to change my sql query to mongo aggregation.
select c.year, c.minor_category, count(c.minor_category) from Crime as c
group by c.year, c.minor_category having c.minor_category = (
Select cc.minor_category from Crime as cc where cc.year=c.year group by
cc.minor_category order by count(*) desc, cc.minor_category limit 1)
I tried do something like this:
db.crimes.aggregate({
$group: {
"_id": {
year: "$year",
minor_category :"$minor_category",
count: {$sum: "$minor_category"}
}
},
},
{
$match : {
minor_category: ?
}
})
But i stuck in $match which is equivalent to having, but i dont know how to make subqueries in mongo like in my sql query.
Can anybody can help me ?

Answer 1

Ok based on the confirmation above , the below query should work.
db.crime.aggregate
([
{"$group":{"_id":{"year":"$year","minor":"$minor"},"count":{"$sum":1}}},
{"$project":{"year":"$_id.year","count":"$count","minor":"$_id.minor","document":"$$ROOT"}},
{"$sort":{"year":1,"count":-1}},
{"$group":{"_id":{"year":"$year"},"orig":{"$first":"$document"}}},
{"$project":{"_id":0,"year":"$orig._id.year","minor":"$orig._id.minor","count":"$orig.count"}}
)]

Answer 2

This translates into the following MongoDB query:
db.crime.aggregate({
$group: { // group by year and minor_catetory
_id: {
"year": "$year",
"minor_category": "$minor_category"
},
"count": { $sum: 1 }, // count all documents per group,
}
}, {
$sort: {
"count": -1, // sort descending by count
"minor_category": 1 // and ascending by minor_category
}
}, {
$group: { // now we get the highst element per year
_id: "$_id.year", // so group by year
"minor_category": { $first: "$_id.minor_category" }, // and get the first (we've sorted the data) value
"count": { $first: "$count" } // same here
}
}, {
$project: { // remove the _id field and add the others in the right order (if needed)
"_id": 0,
"year": "$_id",
"minor_category": "$minor_category",
"count": "$count"
}
})