How to match and group in multiple cases in mongodb aggregation? - mongodb

Question

i have 4 players with there scores in different matches.
e.g
{user: score} -- json keys
{'a': 10}, {'a':12}, {'b':16}
I am trying to find out a way in which i can found sum of single player using aggregation function.
users.aggregation([{$match:{'user':'a'}},{$group:{_id: null, scores:{$sum:'$score'}])
i am repeating same thing for b also and continue
In shot i am doing same thing for different users for too many times.
What is the best way or different way or optimize way, so i can write aggregate query once for all users

Answer 1

You can just match out the required users with the $in clause, and then group as #Sourbh Gupta suggested.
db.users.aggregate([
{$match:{'user':{$in: ['a', 'b', 'c']}}},
{$group:{_id: '$user', scores:{$sum:'$score'}}}
])

Answer 2

group the data on the basis of user. i.e.
users.aggregation([{$group:{_id: "$user", scores:{$sum:'$score'}}}])

Answer 3

Not too sure about your document structures, but if you've got 2 diffrent fields for 2 diffrent scores you can group together and sum then and then project and sum then 2 grouped sums (if that makes sense)
So for example, I have these docuemnts:
> db.scores.find()
{ "_id" : ObjectId("5858ed67b11b12dce194eec8"), "user" : "bob", "score" : { "a" : 10 } }
{ "_id" : ObjectId("5858ed6ab11b12dce194eec9"), "user" : "bob", "score" : { "a" : 12 } }
{ "_id" : ObjectId("5858ed6eb11b12dce194eeca"), "user" : "bob", "score" : { "b" : 16 } }
Notice we have a user bob and he has 2x a scores and 1x b score.
We can now write an aggregation query to do a match for bob then sum the scores.
db.scores.aggregate([
{ $match: { user : "bob" } },
{ $group: { _id : "$user", sumA : { $sum : "$score.a" }, sumB : { $sum : "$score.b" } } },
{ $project: { user: 1, score : { $sum: [ "$sumA", "$sumB" ] } } }
]);
This will give us the following result
{ "_id" : "bob", "score" : 38 }