Return specific field in aggregate - mongodb

Question

i am trying to aggregate the following data:
{
"_id" : ObjectId("527a6b7c24a8874c078b9d10"),
"Name" : "FirstName",
"Link" : "www.mylink.com/123",
"year" : 2013
}
{
"_id" : ObjectId("527a6b7c24a8874c078b9d11"),
"Name" : "FirstName",
"Link" : "www.mylink.com/124",
"year" : 2013
}
{
"_id" : ObjectId("527a6b7c24a8874c078b9d12"),
"Name" : "SecondName",
"Link" : "www.mylink.com/125",
"year" : 2013
}
I want to aggregate number of occurencies of Name field, but also want to return the corresponding Link field in the output of aggregate query. Now I am doing it like this (which does not return the Link field in the output):
db.coll.aggregate([
{ "$match": { "Year": 2013 } },
{ "$group": {
"_id": {
"Name": "$Name"
},
"count": { "$sum": 1 }
}},
{ "$project": {
"_id": "$_id",
"count": 1
}},
{ $sort: {
count: 1
} }
])
The above returns only Name field and count. But how can I also return the corresponding Link field (could be several) in the output of aggregate query?
Best Regards

Answer 1

db.coll.aggregate([
{ "$match": { "year": 2013 } },
{ "$group": {"_id": "$Name", "Link": {$push: "$Link"}, "count": { "$sum": 1 }}},
{ "$project": {"Name": "$_id", _id: 0, "Link": 1, "count": 1}},
{ $sort: {count: 1} }
])
Results in:
{ "Link" : [ "www.mylink.com/125" ], "count" : 1, "Name" : "SecondName" }
{ "Link" : [ "www.mylink.com/123", "www.mylink.com/124" ], "count" : 2, "Name" : "FirstName" }
Ok so the $match was correct except for a typo with 'Year' --> 'year'
The $group could be simplified a little bit. I removed an extra set of brackets so that you get id: 'FirstName' instead of id: { 'name': 'FirstName' } since we can reshape the _id to 'name' in the $project stage anyways.
You needed to add $push or $addToSet to maintain the $Link value in your grouping. $addToSet will allow for unique values in the array only, while $push will add all values, so use whichever at your discretion.
$project and $sort are straightforward, rename and include/exclude whichever fields you would like.