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

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,
.
.
.
.
.

}

Himanshu Sharma · Accepted Answer · 2019-08-28 16:53:11Z

3

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.

edited Aug 28, 2019 at 16:53

answered Aug 28, 2019 at 6:53

Himanshu Sharma

3,0102 gold badges9 silver badges19 bronze badges

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5 Comments

Gopi P Over a year ago

@Sharma exactly that will only give you the exact distinct values. but I need values which are duplicate as well to get the count. ex: {'html':10, 'php': 15} like wise

Himanshu Sharma Over a year ago

@GopiP Updated answer

Gopi P Over a year ago

thanks it works charm. Do we have any even easy to do???? just asking for learning purpose

Himanshu Sharma Over a year ago

@GopiP As per your requirements, we did it that way. If you need a response like this: {"skills":[{"skill":"html":2},{"css":1},{"Java":1}]}, then it can be done easily.

Gopi P Over a year ago

oh thanks @Sharma, If possible can you please share the code too. So, that I could come to know that as well.

Jayshree Rathod · Accepted Answer · 2019-08-28 16:05:50Z

0

You can use aggregates to get the desired results.

db.myCollection.aggregate({ $group: { "skill": "$quizzes.skill","count":{ "$sum":1 }} })

answered Aug 28, 2019 at 16:05

Jayshree Rathod

756 bronze badges

1 Comment

Gopi P Over a year ago

when I run the mentioned query it gives below error The field 'skill' must be an accumulator object

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