Returning only subdocuments in MongoDB - mongodb

Question

I have the following documents in my MongoDB collection:
{
"name": "name",
"items": [
{
"raw": { ... }
"processed": { ... }
},
{
"raw": { ... }
"processed": { ... }
}
]
}
And I'm trying to aggregate / query the database such that I get these items:
[
{"raw": { ... }},
{"raw": { ... }}
]
I'm using the aggregation framework now, but I'm stuck at the part where I want to exclude fields of the outer document.
My current query is:
db.mycollections.aggregate([
{ $unwind: "$items" },
{ $project: { "items.raw": 1 } }
])
And it returns:
[
{"items: {"raw": { ... }}},
{"items: {"raw": { ... }}}
]
Is there a way to only return the subdocuments from the query above?

Answer 1

If you write aggregation with unwind as :
db.mycollections.aggregate({"$unwind":"$items"})
then output looks like :
{ "_id" : ObjectId(), "name" : "name", "items" : { "raw" : {... }, "processed" : { ... } } }
{ "_id" : ObjectId() , "name" : "name", "items" : { "raw" : { ...}, "processed" : { ...} } }
$project passes along the documents with only the specified fields to the next stage in the pipeline. The specified fields can be existing fields from the input documents or newly computed fields.
and you pass $project as your existing fields with items.raw so instead of passing this existing field to project use expression with new field name as raw and changed your aggregation as
db.mycollections.aggregate({"$unwind":"$items"},{"$project":{"raw":"$items.raw"}})
For more details check mongo aggregation pipeline