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How to maintain the top count of array elements in MongoDB?
You can use aggregate framework. Let us first create a collection with documents −
> db.topCountArrayDemo.insertOne(
... {"StudentId":101 , "StudentSubject": ["C", "MongoDB"]}
... );
{
"acknowledged" : true,
"insertedId" : ObjectId("5cc6b3209cb58ca2b005e669")
}
> db.topCountArrayDemo.insertOne(
... {"StudentId":102 , "StudentSubject": ["C", "Java"]}
... );
{
"acknowledged" : true,
"insertedId" : ObjectId("5cc6b3219cb58ca2b005e66a")
}
> db.topCountArrayDemo.insertOne(
... {"StudentId":103 , "StudentSubject": ["C", "MongoDB"]}
... );
{
"acknowledged" : true,
"insertedId" : ObjectId("5cc6b3229cb58ca2b005e66b")
}
Following is the query to display all documents from a collection with the help of find() method −
> db.topCountArrayDemo.find().pretty();
This will produce the following output −
{
"_id" : ObjectId("5cc6b3209cb58ca2b005e669"),
"StudentId" : 101,
"StudentSubject" : [
"C",
"MongoDB"
]
}
{
"_id" : ObjectId("5cc6b3219cb58ca2b005e66a"),
"StudentId" : 102,
"StudentSubject" : [
"C",
"Java"
]
}
{
"_id" : ObjectId("5cc6b3229cb58ca2b005e66b"),
"StudentId" : 103,
"StudentSubject" : [
"C",
"MongoDB"
]
}
Following is the query to maintain the top count of array elements in MongoDB −
> db.topCountArrayDemo.aggregate(
... [
... {
... $unwind: "$StudentSubject"
... },
... {
... $group: {
... _id: "$StudentSubject",
... Frequency: {$sum: 1}
... }
... },
... {
... $sort: {Frequency:-1}
... },
... {
... $limit: 2
... }
... ]
... ).pretty();
This will produce the following output −
{ "_id" : "C", "Frequency" : 3 }
{ "_id" : "MongoDB", "Frequency" : 2 }
Updated on: 2019-07-30T22:30:26+05:30
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