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 }