You can use $pull operator for this. Let us first create a collection with documents. Here, we have also added an empty object −> db.removeEmptyObjectsDemo.insertOne(    {       "_id" :101,       "LoginDate" :new ISODate(),       "UserDetails" : [          {             "UserName" : "John"          },          {          },          {             "UserName" : "Sam"          }       ]    } ); { ... Read More

To return documents of a collection without objectId, set _id:0. Let us first create a collection with documents −> db.returnDocumentWithoutObjectId.insertOne({"Name":"Carol", "Age":25}); {    "acknowledged" : true,    "insertedId" : ObjectId("5ce8ba6c78f00858fb12e8fa") } > db.returnDocumentWithoutObjectId.insertOne({"Name":"Sam", "Age":21}); {    "acknowledged" : true,    "insertedId" : ObjectId("5ce8ba6d78f00858fb12e8fb") } > db.returnDocumentWithoutObjectId.insertOne({"Name":"John", "Age":23}); {    "acknowledged" : true,    "insertedId" : ObjectId("5ce8ba6f78f00858fb12e8fc") }Following is the query to display all documents from a collection with the help of find() method −> db.returnDocumentWithoutObjectId.find();This will produce the following output −{ "_id" : ObjectId("5ce8ba6c78f00858fb12e8fa"), "Name" : "Carol", "Age" : 25 } { "_id" : ObjectId("5ce8ba6d78f00858fb12e8fb"), "Name" : "Sam", "Age" : ... Read More

Yes, to display a database in the list, first create a database and add collection(s), else it won’t be visible in the list. After that, use the SHOW dbs command to display the database name in the list of databases.Following is the query to create a database −> use webcustomertracker; switched to db webcustomertrackerLet us first create a collection with documents −> db.first_Collection.insert({"Name":"Chris"}); WriteResult({ "nInserted" : 1 })Following is the query to display all documents from a collection with the help of find() method −> db.first_Collection.find();This will produce the following output −{ "_id" : ObjectId("5ce2760836e8b255a5eee94a"), "Name" : "Chris" }Following is ... Read More

You can’t directly update _id field i.e. write some script to update. Let us first create a collection with documents −> db.updatingIdFieldDemo.insertOne({"StudentName":"Chris"}); {    "acknowledged" : true,    "insertedId" : ObjectId("5ce271bb36e8b255a5eee949") }Following is the query to display all documents from a collection with the help of find() method −> db.updatingIdFieldDemo.find();This will produce the following output −{ "_id" : ObjectId("5ce271bb36e8b255a5eee949"), "StudentName" : "Chris" }Following is the query to update _id field in MongoDB −> var myDocument=db.updatingIdFieldDemo.findOne({StudentName:"Chris"}); > myDocument._id = 101; 101 > db.updatingIdFieldDemo.save(myDocument); WriteResult({ "nMatched" : 0, "nUpserted" : 1, "nModified" : 0, "_id" : 101 }) > db.updatingIdFieldDemo.remove({_id:ObjectId("5ce271bb36e8b255a5eee949")}); WriteResult({ ... Read More

To reduce the time to find record in MongoDB, you can use index. Following is the syntax −db.yourCollectionName.createIndex({yourFieldName:1});You can follow the below approaches to create index for field names based on number, text, hash, etc.First ApproachLet us create an index. Following is the query −> db.takeLessTimeToSearchDemo.createIndex({"EmployeeName":1}); {    "createdCollectionAutomatically" : true,    "numIndexesBefore" : 1,    "numIndexesAfter" : 2,    "ok" : 1 }Second ApproachTo understand the above concept, let us create another index −> db.takeLessTimeToSearchDemo1.createIndex({"EmployeeName":"text"}); {    "createdCollectionAutomatically" : true,    "numIndexesBefore" : 1,    "numIndexesAfter" : 2,    "ok" : 1 }Third ApproachLet us now create another index −> db.takeLessTimeToSearchDemo2.createIndex({"EmployeeName":"hashed"}); {    "createdCollectionAutomatically" : true,    "numIndexesBefore" : 1,    "numIndexesAfter" : 2,    "ok" : 1 }

Use aggregate framework along with $literal operator. Let us first create a collection with documents −> db.replaceValueDemo.insertOne(    {       _id : 100,       "EmployeeName" :"Chris",       "EmployeeOtherDetails": {          "EmployeeDesignation" : "HR",          "EmployeeAge":27       }    } ); { "acknowledged" : true, "insertedId" : 100 } > db.replaceValueDemo.insertOne(    {       _id : 101,       "EmployeeName" :"David",       "EmployeeOtherDetails": {          "EmployeeDesignation" : "Tester",          "EmployeeAge":26       }    } ... Read More

You can use $in operator for this. Let us first create a collection with documents −> db.searchForStringOrNumberDemo.insertOne(    {       "_id": new ObjectId(),       "StudentName": "Larry",       "StudentDetails": {          "StudentMarks": {             "StudentMongoDBMarks": [44]          }       }    } ); {    "acknowledged" : true,    "insertedId" : ObjectId("5ce2407c36e8b255a5eee944") } > db.searchForStringOrNumberDemo.insertOne( { "_id": new ObjectId(), "StudentName": "Larry", "StudentDetails": { "StudentMarks": { "StudentMongoDBMarks": ["44"] } } } ); {    "acknowledged" : true,    "insertedId" : ObjectId("5ce240f036e8b255a5eee945") }Following is ... Read More

To return specific fields, use aggregate $project. Let us first create a collection with documents −> db.returnSpecificFieldDemo.insertOne(    {       "StudentId":1,       "StudentDetails": [          {             "StudentName":"Larry",             "StudentAge":21,             "StudentCountryName":"US"          },          {             "StudentName":"Chris",             "StudentAge":23,             "StudentCountryName":"AUS"          }       ]    } ); {    "acknowledged" ... Read More

For this, use the aggregate framework. Let us first create a collection with documents −> db.aggregateArrayDemo.insertOne(    {       "_id":100,       "UserDetails": [          {             "UserName": "John",             "UserLoginYear":2010          },          {             "UserName": "Carol",             "UserLoginYear":2019          }       ]    } ); { "acknowledged" : true, "insertedId" : 100 }Following is the query to display all documents from a collection with the help of find() method −> db.aggregateArrayDemo.find().pretty();This will produce the following output −{    "_id" : 100,    "UserDetails" : [       {          "UserName" : "John",          "UserLoginYear" : 2010       },       {          "UserName" : "Carol",          "UserLoginYear" : 2019       }    ] }Following is the query to aggregate array documents −> db.aggregateArrayDemo.aggregate([    {       $match: { _id: 100 }    },    {       $project: {          Minimum: { $min: "$UserDetails.UserLoginYear" },          Maximum: { $max: "$UserDetails.UserLoginYear" }       }    } ]);This will produce the following output −{ "_id" : 100, "Minimum" : 2010, "Maximum" : 2019 }

To get the matching document, use $elemMatch. Let us first create a collection with documents −> db.getMatchingDocumentDemo.insertOne(    {       _id :1,       "UserDetails":[          {             "UserName":"John",             "UserAge":23          }       ]    } ); { "acknowledged" : true, "insertedId" : 1 } > db.getMatchingDocumentDemo.insertOne( { _id :2, "UserDetails":[ { "UserName":"Larry", "UserAge":24 } ] } ); { "acknowledged" : true, "insertedId" : 2 }Following is the query to display all documents from a collection with the ... Read More