get only a particular field from embedded array object in mongodb collection - mongodb

Question

I'm having a MongoDB collection which looks same as below document and I wanted to find only one field values count which presents within the embedded array object.
I tried below query to fetch the data but doesn't work
db.mycollection.find({'quizzes':{skill:'html'}}).pretty()
below is the mongo document structure with sample value. the structure is the same as my original document
{
"user": "values",
"date": "234-234-234-234",
"quizzes":[
{
"skill": "html",
"score": "12"
}
]
}
From the above document, I wanted to fetch only the skill field values which present within quizzes array which is an embedded document.
my output should be like
{
"html": 10,
"php": 20,
"C#": 15,
"java": 18,
.
.
.
.
.
}

Answer 1

You can use the distinct() method. The following query can get us the expected output:
db.mycollection.distinct("quizzes.skill");
For more information, please check https://docs.mongodb.com/manual/reference/method/db.collection.distinct/
Edit I: Calculating count of skills too
db.collection.aggregate([
{
$unwind:"$quizzes"
},
{
$group:{
"_id":"$quizzes.skill",
"k":{
$first:"$quizzes.skill"
},
"v":{
$sum:1
}
}
},
{
$project:{
"_id":0
}
},
{
$group:{
"_id":null,
"data":{
$push:"$$ROOT"
}
}
},
{
$project:{
"data":{
$arrayToObject:"$data"
}
}
},
{
$replaceRoot:{
"newRoot":"$data"
}
}
]).pretty()
Data set:
{
"_id" : ObjectId("5d66ad357d0ab652c42315f7"),
"user" : "values",
"date" : "234-234-234-234",
"quizzes" : [
{
"skill" : "html",
"score" : "12"
},
{
"skill" : "css",
"score" : "10"
}
]
}
{
"_id" : ObjectId("5d66ad357d0ab652c42315f8"),
"user" : "values2",
"date" : "234-234-234-234",
"quizzes" : [
{
"skill" : "Java",
"score" : "12"
},
{
"skill" : "html",
"score" : "10"
}
]
}
Output:
{ "Java" : 1, "css" : 1, "html" : 2 }
Explanation: We are creating a key and value pair (k,v) where 'k' is the skill and 'v' is the count of skill occurrence. The reason behind taking field names as 'k' and 'v' is because $arrayToObject only takes these fields only. Later on, all keys and values are merged to prepare the final document.

Answer 2

You can use aggregates to get the desired results.
db.myCollection.aggregate({ $group: { "skill": "$quizzes.skill","count":{ "$sum":1 }} })