MongoDB order by a sum on a subset - mongodb

Question

I have the following collection:
error_reports
[
{
"_id":{
"$oid":"5184de1261"
},
"date":"29/04/2013",
"errors":[
{
"_id":"10",
"failures":2,
"alerts":1,
},
{
"_id":"11",
"failures":7,
"alerts":4,
}
]
},
{
"_id":{
"$oid":"5184de1262"
},
"date":"30/04/2013",
"errors":[
{
"_id":"15",
"failures":3,
"alerts":2,
},
{
"_id":"16",
"failures":9,
"alerts":1,
}
]
}
]
Is it possible to retrieve the list of documents with failures and alerts sum sorted by failures in descending order? I am new to mongodb, I have been searching for 2 days but I can't figure out what is the proper query...
I tried something like this :
db.error_reports.aggregate(
{ $sort : { failures: -1} },
{ $group:
{ _id: "$_id",
failures: { "$sum": "$errors.failures" }
}
}
);
But it didn't work, I think it is because of the $sum: $errors.failures thing, I would like to sum this attribute on every item of the day_hours subcollection but I don't know of to do this in a query...

Answer 1

You were very close with your attempt. The only thing missing is the $unwind aggregation operator. $unwind basically splits each document out based on a sub-document. So before you group the failures and alerts, you unwind the errors, like so:
db.error_reports.aggregate(
{ $unwind : '$errors' },
{ $group : {
_id : '$_id',
'failures' : { $sum : '$errors.failures' },
'alerts' : { $sum : '$errors.alerts' }
} },
{ $sort : { 'failures': -1 } }
);
Which gives you the follow result:
{
"result" : [
{
"_id" : ObjectId("5184de1262"),
"failures" : 12,
"alerts" : 3
},
{
"_id" : ObjectId("5184de1261"),
"failures" : 9,
"alerts" : 5
}
],
"ok" : 1
}