Mongodb accessing documents - mongodb

Question

I've the following db:
{ "_id" : 1, "results" : [ { "product" : "abc", "score" : 10 }, { "product" : "xyz", "score" : 5 } ] }
{ "_id" : 2, "results" : [ { "product" : "abc", "score" : 8 }, { "product" : "xyz", "score" : 7 } ] }
{ "_id" : 3, "results" : [ { "product" : "abc", "score" : 7 }, { "product" : "xyz", "score" : 8 } ] }
I want to show the first score of each _id, i tried the following:
db.students.find({},{"results.$":1})
But it doesn't seem to work, any advice?

Answer 1

You can take advantage of aggregation pipeline to solve this.
Use $project in conjunction with $arrayElemAt to point to appropriate node index in the array.
So, to extract the documents of the first score, have written below query.
db.students.aggregate([ {$project: { scoredoc:{$arrayElemAt:["$results",0]}} } ]);
In case if you just wish to have scores excluding product, use $results.score as shown below.
db.students.aggregate([ {$project: { scoredoc:{$arrayElemAt:["$results.score",0]}} } ]);
Here scoredoc object will have all documents of first score element.
Hope this helps!

Answer 2

According to above mentioned description please try executing following query in MongoDB shell
db.students.find(
{results:
{$elemMatch:{score:{$exists:true}}}}, {'results.$.score':1}
)
According to MongoDB documentation
The positional $ operator limits the contents of an from the
query results to contain only the first element matching the query
document.
Hence in above mentioned query positional $ operator is used in projection section to retrieve first score of each document.