To find maximum score, use GROUP() to group documents in a collection. Let us create a collection with documents −> db.demo114.insertOne({"Score":60, "Name":"Chris"}); {    "acknowledged" : true,    "insertedId" : ObjectId("5e2efc57d8f64a552dae6354") } > db.demo114.insertOne({"Score":87, "Nam+e":"Chris"}); {    "acknowledged" : true,    "insertedId" : ObjectId("5e2efc5ad8f64a552dae6355") } > db.demo114.insertOne({"Score":45, "Name":"David"}); {   ... Read More

To create views in MongoDB, use createView(). Let us create a collection with documents −> db.demo113.insertOne( ... { _id: 1, StudentId: "101", "Details": { Name: "Chris", Age: 21 }, Subject: "MySQL" } ... ); { "acknowledged" : true, "insertedId" : 1 }Display all documents from a collection with the help ... Read More

To update, you need to use update() for the field whose collection is set with capped: true. Let us create a collection with documents −> db.createCollection("Demo112", { capped : true, size : 14, max : 3 } ); { "ok" : 1 } > db.demo112.insertOne({"Name":"Chris"}); {    "acknowledged" : true, ... Read More

For this, use aggregate() in MongoDB. In that, use $slice to slice array inside array. Let us create a collection with documents −> db.demo111.insertOne( ...    { ...       "_id" : 101, ...       "Name" : "Chris", ...       "Details" : [ ...   ... Read More

To update data in MongoDB, use update(). Let us create a collection with documents −> db.demo110.insertOne({"Name":"Chris"}); {    "acknowledged" : true,    "insertedId" : ObjectId("5e2eeb949fd5fd66da21447b") } > db.demo110.insertOne({"Name":"David"}); {    "acknowledged" : true,    "insertedId" : ObjectId("5e2eeb9f9fd5fd66da21447c") } > db.demo110.insertOne({"Name":"Bob"}); {    "acknowledged" : true,    "insertedId" : ObjectId("5e2eeba39fd5fd66da21447d") }Display ... Read More

To get the data of the nested JSON object in MongoDB, use findOne(). Let us create a collection with documents −> db.demo109.insertOne( ...    { ...       "Name" : "Chris", ...       "Subjects" : [ ...          { ...         ... Read More

To remove duplicate entries by two keys, use aggregate() in MongoDB. Let us create a collection with documents −> db.demo108.insertOne({"Value1":23, "Value2":24}); {    "acknowledged" : true,    "insertedId" : ObjectId("5e2ee3e49fd5fd66da214477") } > db.demo108.insertOne({"Value1":23, "Value2":25}); {    "acknowledged" : true,    "insertedId" : ObjectId("5e2ee3f29fd5fd66da214478") } > db.demo108.insertOne({"Value1":23, "Value2":24}); {    "acknowledged" ... Read More

To use variables, work with var in MongoDB. Let us create a collection with documents −> db.demo107.insertOne({"Name":"Chris"}); {    "acknowledged" : true,    "insertedId" : ObjectId("5e2ee1b19fd5fd66da214471") } > db.demo107.insertOne({"Name":"Bob"}); {    "acknowledged" : true,    "insertedId" : ObjectId("5e2ee1b49fd5fd66da214472") } > db.demo107.insertOne({"Name":"David"}); {    "acknowledged" : true,    "insertedId" : ObjectId("5e2ee1b89fd5fd66da214473") ... Read More

You can use bulkWrite(). Let us create a collection with documents −> db.demo105.insertOne( { _id:'101', Name:'Chris', Details:[{ Marks1:60, Marks2:70, Marks3:70 }, { Marks1:70, Marks2:70, Marks3:90 }] } ); { "acknowledged" : true, "insertedId" : "101" }Display all documents from a collection with the help of find() method −> db.demo105.find().pretty();This will ... Read More

To implement $dateToString on array items, use aggregate() in MongoDB. Let us create a collection with documents −> db.demo104.insertOne( ...    { ... ...       "AppName" : "Online Book", ...       "Details" : [ ...          { ...           ... Read More