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MongoDB query for ranking / search count?
For this, use aggregate() in MongoDB. Let us create a collection with documents −
> db.demo120.insertOne(
... {
... 'Name': 'Chris',
... 'Subjects': [ 'MySQL', 'MongoDB', 'Java', 'Python' ]
... }
... );
{
"acknowledged" : true,
"insertedId" : ObjectId("5e2f11aed8f64a552dae6365")
}
> db.demo120.insertOne(
... {
... 'Name': 'Bob',
... 'Subjects': [ 'C', 'MongoDB' ]
... }
... );
{
"acknowledged" : true,
"insertedId" : ObjectId("5e2f11afd8f64a552dae6366")
}Display all documents from a collection with the help of find() method −
> db.demo120.find();
This will produce the following output −
{ "_id" : ObjectId("5e2f11aed8f64a552dae6365"), "Name" : "Chris", "Subjects" : [ "MySQL", "MongoDB", "Java", "Python" ] }
{ "_id" : ObjectId("5e2f11afd8f64a552dae6366"), "Name" : "Bob", "Subjects" : [ "C", "MongoDB" ] }Following is the query for MongoDB ranking / search count −
> var s = ['MySQL', 'Java', 'MongoDB'];
> db.demo120.aggregate([
... { "$match": { "Subjects": { "$in": s } } },
... {
... "$addFields": {
... "RankSearch": {
... "$divide": [
... { "$size": { "$setIntersection": ["$Subjects",s] } },
... { "$size": "$Subjects" }
... ]
... }
... }
... },
... { "$sort": { "RankSearch": -1 } }
... ])This will produce the following output −
{ "_id" : ObjectId("5e2f11aed8f64a552dae6365"), "Name" : "Chris", "Subjects" : [ "MySQL", "MongoDB", "Java", "Python" ], "RankSearch" : 0.75 }
{ "_id" : ObjectId("5e2f11afd8f64a552dae6366"), "Name" : "Bob", "Subjects" : [ "C", "MongoDB" ], "RankSearch" : 0.5 }
Updated on: 31-Mar-2020
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