Found 1359 Articles for MongoDB

You can use distinct() method in MongoDB to get distinct record values. The syntax is as follows −db.yourCollectionName.distinct(“yourFieldName”);To understand the above syntax, let us create a collection with document. The query to create a collection with document is as follows −> db.distinctRecordDemo.insertOne({"StudentId":1, "StudentName":"John", "StudentAge":21}); {    "acknowledged" : true,    "insertedId" : ObjectId("5c77a78299b97a65744c1b50") } > db.distinctRecordDemo.insertOne({"StudentId":2, "StudentName":"John", "StudentAge":22}); {    "acknowledged" : true,    "insertedId" : ObjectId("5c77a78b99b97a65744c1b51") } > db.distinctRecordDemo.insertOne({"StudentId":3, "StudentName":"Carol", "StudentAge":21}); {    "acknowledged" : true,    "insertedId" : ObjectId("5c77a79a99b97a65744c1b52") } > db.distinctRecordDemo.insertOne({"StudentId":4, "StudentName":"Carol", "StudentAge":26}); {    "acknowledged" : true,    "insertedId" : ObjectId("5c77a7a499b97a65744c1b53") } > db.distinctRecordDemo.insertOne({"StudentId":5, "StudentName":"Sam", "StudentAge":24}); ... Read More

You can achieve this with the help of an aggregate framework. To understand the concept, let us create a collection with the document. The query to create a collection with a document is as follows −> db.countGroupByDemo.insertOne({"StudentId":10, "StudentName":"John"}); {    "acknowledged" : true,    "insertedId" : ObjectId("5c7700871e9c5dd6f1f78296") } > db.countGroupByDemo.insertOne({"StudentId":10, "StudentName":"Carol"}); {    "acknowledged" : true,    "insertedId" : ObjectId("5c77008f1e9c5dd6f1f78297") } > db.countGroupByDemo.insertOne({"StudentId":20, "StudentName":"Sam"}); {    "acknowledged" : true,    "insertedId" : ObjectId("5c7700971e9c5dd6f1f78298") } > db.countGroupByDemo.insertOne({"StudentId":30, "StudentName":"Mike"}); {    "acknowledged" : true,    "insertedId" : ObjectId("5c7700a21e9c5dd6f1f78299") } > db.countGroupByDemo.insertOne({"StudentId":30, "StudentName":"David"}); {    "acknowledged" : true,    "insertedId" : ObjectId("5c7700aa1e9c5dd6f1f7829a") } ... Read More

You can use $ne(Not Equal) operator for this. To understand the concept, let us create a collection with document. The query to create a collection with document is as follows −> db.arrayFieldIsNotEmptyDemo.insertOne({"StudentName":"Larry", "StudentTechnicalSubject":["Java", "C"]}); {    "acknowledged" : true,    "insertedId" : ObjectId("5c76fe2f1e9c5dd6f1f78291") } > db.arrayFieldIsNotEmptyDemo.insertOne({"StudentName":"Mike", "StudentTechnicalSubject":[]}); {    "acknowledged" : true,    "insertedId" : ObjectId("5c76fe3b1e9c5dd6f1f78292") } > db.arrayFieldIsNotEmptyDemo.insertOne({"StudentName":"Sam", "StudentTechnicalSubject":["MongoDB"]}); {    "acknowledged" : true,    "insertedId" : ObjectId("5c76fe491e9c5dd6f1f78293") } > db.arrayFieldIsNotEmptyDemo.insertOne({"StudentName":"Carol", "StudentTechnicalSubject":[]}); {    "acknowledged" : true,    "insertedId" : ObjectId("5c76fe521e9c5dd6f1f78294") } > db.arrayFieldIsNotEmptyDemo.insertOne({"StudentName":"David", "StudentTechnicalSubject":["MySQL", "SQL Server"]}); {    "acknowledged" : true,    "insertedId" : ObjectId("5c76fe661e9c5dd6f1f78295") }Display all documents ... Read More

You can use the $exists and $ne operator for this. To understand the concept further, let us create a collection with document. The query to create a collection with document is as follows −> db.checkFieldExistDemo.insertOne({"EmployeeId":1, "EmployeeName":"John", "isMarried":true, "EmployeeSalary":4648585}); {    "acknowledged" : true,    "insertedId" : ObjectId("5c76f7b31e9c5dd6f1f78281") } > db.checkFieldExistDemo.insertOne({"StudentId":2, "StudentName":"John", "isMarried":false, " StudentAge":19}); {    "acknowledged" : true, 0    "insertedId" : ObjectId("5c76f7e11e9c5dd6f1f78282") }Display all documents from a collection with the help of find() method. The query is as follows −> db.checkFieldExistDemo.find().pretty();Output{    "_id" : ObjectId("5c76f7b31e9c5dd6f1f78281"),    "EmployeeId" : 1,    "EmployeeName" : "John",    "isMarried" : true,   ... Read More

You can get documents with max attribute per group in a collection using $sort operator along with $group statement.To understand the concept further, let us create a collection with document. The query to create a collection with document is as follows −> db.maxAttributePerGroup.insertOne({"StudentFirstName":"John", "StudentLastName":"Smith    ", "StudentAge":29, "StudentId":10}); {    "acknowledged" : true,    "insertedId" : ObjectId("5c76ee341e9c5dd6f1f78277") } > db.maxAttributePerGroup.insertOne({"StudentFirstName":"Carol", "StudentLastName":"Taylo    r", "StudentAge":19, "StudentId":10}); {    "acknowledged" : true,    "insertedId" : ObjectId("5c76ee4e1e9c5dd6f1f78278") } > db.maxAttributePerGroup.insertOne({"StudentFirstName":"Adam", "StudentLastName":"Smit    h", "StudentAge":34, "StudentId":20}); {    "acknowledged" : true,    "insertedId" : ObjectId("5c76ee631e9c5dd6f1f78279") } > db.maxAttributePerGroup.insertOne({"StudentFirstName":"Bob", "StudentLastName":"Taylor"    , "StudentAge":58, "StudentId":20}); { ... Read More

In order to replace substring in MongoDB document, you can use the replace() function. To understand it further, let us create a collection with document. The query to create a collection with document is as follows −> db.replaceSubstringDemo.insertOne({"WebsiteURL":"www.gogle.com"}); {    "acknowledged" : true,    "insertedId" : ObjectId("5c76eaf21e9c5dd6f1f78276") }Display all documents from a collection with the help of find() method. The query is as follows −> db.replaceSubstringDemo.find().pretty();Output{    "_id" : ObjectId("5c76eaf21e9c5dd6f1f78276"),    "WebsiteURL" : "www.gogle.com" }Here is the query to replace substring in MongoDB document −> db.replaceSubstringDemo.find({WebsiteURL:"www.gogle.com"}).forEach(function(url, k){    ... url.WebsiteURL=url.WebsiteURL.replace("www.gogle.com", "www.google.com");    ... db.replaceSubstringDemo.save(url)    ... });Let us display the ... Read More

You can use $facet operator for this. To understand the concept, let us create a collection with document. The query to create a collection with document is as follows −> db.totalDocumentDemo.insertOne({"InstructorId":100, "InstructorName":"Larry", "InstructorFav ouriteSubject":["Java", "MongoDB", "Python"]}); {    "acknowledged" : true,    "insertedId" : ObjectId("5c76e6701e9c5dd6f1f78274") } > db.totalDocumentDemo.insertOne({"InstructorId":200, "InstructorName":"Sam", "InstructorFav ouriteSubject":["SQL Server", "C#", "Javascript"]}); {    "acknowledged" : true,    "insertedId" : ObjectId("5c76e69c1e9c5dd6f1f78275") }Display all documents from a collection with the help of find() method. The query is as follows −> db.totalDocumentDemo.find().pretty();Output{    "_id" : ObjectId("5c76e6701e9c5dd6f1f78274"),    "InstructorId" : 100,    "InstructorName" : "Larry",    "InstructorFavouriteSubject" : [     ... Read More

To perform Regex search on integer value, you need to use $where operator. The syntax is as follows:db.yourCollectionName.find({ $where: "/^yourIntegerPatternValue.*/.test(this.yourFieldName)" });To understand the above concept, let us create a collection with document. The query to create a collection with document is as follows:> db.regExpOnIntegerDemo.insertOne({"StudentId":2341234}); {    "acknowledged" : true,    "insertedId" : ObjectId("5c70370c75eb1743ddddce21") } > db.regExpOnIntegerDemo.insertOne({"StudentId":123234}); {    "acknowledged" : true,    "insertedId" : ObjectId("5c70371175eb1743ddddce22") } > db.regExpOnIntegerDemo.insertOne({"StudentId":9871234}); {    "acknowledged" : true,    "insertedId" : ObjectId("5c70371875eb1743ddddce23") } > db.regExpOnIntegerDemo.insertOne({"StudentId":2345612}); {    "acknowledged" : true,    "insertedId" : ObjectId("5c70372275eb1743ddddce24") } > db.regExpOnIntegerDemo.insertOne({"StudentId":1239812345}); {    "acknowledged" : true,    "insertedId" : ... Read More

You can achieve this with the help of aggregate framework in MongoDB. To understand it, let us create a collection with document. The query to create a collection with document is as follows:> db.sortInnerArrayDemo.insertOne( ... ...    { ...       "EmployeeDetails": ...       { ...          "EmployeeAddress": ...          { ...             "EmployeeCountry": ...             [ ...                { ...                   "EmployeeZipCode":1003, ...       ... Read More

You can use the $size operator for internal array size in MongoDB. The syntax is as follows:db.internalArraySizeDemo.aggregate(    [       {          $group: {             _id:yourObjectIdValue,             anyFieldName: {$first: {$size: "$yourArrayName" }}          }       }    ] );To understand the above syntax, let us create a collection with some documents. The query to create a collection with documents are as follows:>db.internalArraySizeDemo.insertOne({"EmployeeName":"Mike", "EmployeeTechnology":["Jav a Web Development", "Python Web Development"]}); {    "acknowledged" : true,    "insertedId" : ObjectId("5c6eff586fd07954a48906b2") } > ... Read More