aggregate request MongoDB - mongodb

Question

I'd like get a multiple fields in a collection list with a condition. I tried a aggregate request but i have an error.
My request
db.people.aggregate({$match:{createdDate:{$exists:true},"ad":"noc2"}},{$group:{value2:$value2}});
My Json :
db.test.findOne();
{
"_id" : ObjectId("51e7dd16d2f8db27b56ea282"),
"ad" : "noc2",
"list" : {
"p45" : {
"id" : "p45",
"date" : ISODate("2014-01-01T12:18:30.568Z"),
"value3" : 21,
"value1" : 100,
"value2" : 489
},
"p6" : {
"id" : "p6"
"date" : ISODate("2013-07-18T12:18:30.568Z"),
"value3" : 21,
"value1" : 100,
"value2" : 489
},
"p4578" : {
"id" : "4578"
"date" : ISODate("2013-07-18T12:18:30.568Z"),
"value3" : 21,
"value1" : 100,
"value2" : 489
}
}
}
I want to get this json, for example, in result :
{id:p45,value:587},{id:p4578,value:47},{id:p6,value:2}

Answer 1

There are several issues with your sample document and aggregation:
the sample doc will not match your aggregation query because you are matching on createdDate field existing
the $group() operator works on a document level and needs an _id field to group by
your list field is an embedded document, not an array
aside from formatting, there is no obvious way to relate the sample values to the calculated result you are looking for
Here is an adjusted example document with the list as an array as well as some unique values for each item that happen to add up to the value numbers you mentioned:
db.people.insert({
"ad" : "noc2",
"createdDate" : ISODate(),
"list" : [
{
"id" : "p45",
"date" : ISODate("2014-01-01T12:18:30.568Z"),
"value3" : 21,
"value1" : 77,
"value2" : 489
},
{
"id" : "p6",
"date" : ISODate("2013-07-18T12:18:30.568Z"),
"value3" : 20,
"value1" : 20,
"value2" : 7
},
{
"id" : "4578",
"date" : ISODate("2013-07-18T12:18:30.568Z"),
"value3" : 21,
"value1" : 300,
"value2" : -319
}
]
})
You can now use the $unwind operator to extract the matching subdocuments.
It is unclear from the current question description what $group operation you are trying to achieve and whether you actually need $group.
Here is an example using the Aggregation Framework to $add (total) the three values for each list item:
db.people.aggregate(
// Find relevant docs (can use index if available)
{ $match: {
createdDate: { $exists:true },
"ad":"noc2"
}},
// Convert list array to stream of documents
{ $unwind: "$list" },
// Add (value1, value2, value3) for each list item
{ $project: {
_id: "$list.id",
value: { $add: [ "$list.value1", "$list.value2", "$list.value3"] }
}}
);
Sample output:
{
"result" : [
{
"_id" : "p45",
"value" : 587
},
{
"_id" : "p6",
"value" : 47
},
{
"_id" : "4578",
"value" : 2
}
],
"ok" : 1
}
Note that I've only aggregated using a single document for this example, and the output will be one result document for every item in the list array.
In real usage you may want to add an additional aggregation step to $group by document _id or some other criteria, depending on the outcome you are trying to achieve.