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Big Data Analytics Articles - Page 79 of 167
 
 
			
			555 Views
Let us create a collection with documents −> db.demo345.insertOne({ ... "UserName" : "Robert", ... "UserDetails" : [ ... { ... "isMarried" : false, ... "CountryName":"US" ... ... }, ... { ... "isMarried" : true, ... "CountryName":"UK" ... ... }, ... { ... "isMarried" : false, ... ... Read More
 
 
			
			156 Views
Let us create a collection with documents −> db.demo344.insertOne({"startDate":"2020-02-24 10:50:00", "endDate":"2020-02-24 11:50:00"}); { "acknowledged" : true, "insertedId" : ObjectId("5e53f52cf8647eb59e5620aa") } > db.demo344.insertOne({"startDate":"2020-02-24 08:00:00", "endDate":"2020-02-24 11:50:50"}); { "acknowledged" : true, "insertedId" : ObjectId("5e53f53df8647eb59e5620ab") }Display all documents from a collection with the help of find() method −> db.demo344.find();This will produce the following output −{ "_id" : ObjectId("5e53f52cf8647eb59e5620aa"), "startDate" : "2020-02-24 10:50:00", "endDate" : "2020-02-24 11:50:00" } { "_id" : ObjectId("5e53f53df8647eb59e5620ab"), "startDate" : "2020-02-24 08:00:00", "endDate" : "2020-02-24 11:50:50" }Following is how to query objects with the longest time period −> db.demo344.aggregate([ ... { $addFields: { ... ... Read More
 
 
			
			99 Views
For such conversion, use aggregate. Let us create a collection with documents −> db.demo343.insertOne({ ... _id: 101, ... UserName: "Chris", ... details: [ ... {"Name":"John"}, ... {"Name":"David"} ... ] ... } ... ); { "acknowledged" : true, "insertedId" : 101 }Display all documents from a collection with the help of find() method −> db.demo343.find().pretty();This will produce the following output −{ "_id" : 101, "UserName" : "Chris", "details" : [ { "Name" : "John" }, ... Read More
 
 
			
			835 Views
To get only specific fields in nested array documents, use $filter along with $project. Let us create a collection with documents −> db.demo342.insertOne({ ... "Id": "101", ... "details1" : { ... "details2" : [ ... { ... "details3" : [ ... { ... "Name": "Mike", ... "CountryName" : "US" ... ... Read More
 
 
			
			473 Views
To find value in an array within a range, use $gt and $lt. Let us create a collection with documents −> db.demo341.insertOne({ ... "Name": "Chris", ... "productDetails" : [ ... { ... "ProductPrice" : { ... "Price" : 800 ... } ... }, ... { ... "ProductPrice" : { ... "Price" : 400 ... } ... ... Read More
 
 
			
			243 Views
To get only a single result, use findOne() and fetch on the basis of id. Let us create a collection with documents −> db.demo340.insertOne({_id:1, "Name":"Chris", Age:21}); { "acknowledged" : true, "insertedId" : 1 } > db.demo340.insertOne({_id:2, "Name":"David", Age:23}); { "acknowledged" : true, "insertedId" : 2 } > db.demo340.insertOne({_id:3, "Name":"Bob", Age:20}); { "acknowledged" : true, "insertedId" : 3 } > db.demo340.insertOne({_id:4, "Name":"Sam", Age:19}); { "acknowledged" : true, "insertedId" : 4 }Display all documents from a collection with the help of find() method −> db.demo340.find();This will produce the following output −{ "_id" : 1, "Name" : "Chris", "Age" : 21 } { ... Read More
 
 
			
			242 Views
For this, use aggregate along with $zip. The zip is used to transpose an array. Let us create a collection with documents −> db.demo339.insertOne({Id:101, Score1:["98", "56"], Score2:[67, 89]}); { "acknowledged" : true, "insertedId" : ObjectId("5e529ee5f8647eb59e5620a2") }Display all documents from a collection with the help of find() method −> db.demo339.find();This will produce the following output −{ "_id" : ObjectId("5e529ee5f8647eb59e5620a2"), "Id" : 101, "Score1" : [ "98", "56" ], "Score2" : [ 67, 89 ] }Following is the query to zip two arrays with $zip and create a new array of object −> db.demo339.aggregate([ ... { ... ... Read More
 
 
			
			385 Views
The $concatArrays is used to concatenate arrays to return the concatenated array.Let us create a collection with documents −> db.demo338.insertOne({"Name":"Chris", "Marks1":[ [56, 67, 45], [67, 89, 90, 91]]}); { "acknowledged" : true, "insertedId" : ObjectId("5e5299baf8647eb59e56209f") }Display all documents from a collection with the help of find() method −> db.demo338.find().pretty();This will produce the following output −{ "_id" : ObjectId("5e5299baf8647eb59e56209f"), "Name" : "Chris", "Marks1" : [ [ 56, 67, 45 ], [ ... Read More
 
 
			
			2K+ Views
To get the sum, use $sum along with aggregate(). Let us create a collection with documents −> db.demo337.insertOne({"Amount":100}); { "acknowledged" : true, "insertedId" : ObjectId("5e5231e5f8647eb59e56209b") } > db.demo337.insertOne({"Amount":500}); { "acknowledged" : true, "insertedId" : ObjectId("5e5231e9f8647eb59e56209c") } > db.demo337.insertOne({"Amount":400}); { "acknowledged" : true, "insertedId" : ObjectId("5e5231ebf8647eb59e56209d") }Display all documents from a collection with the help of find() method −> db.demo337.find();This will produce the following output −{ "_id" : ObjectId("5e5231e5f8647eb59e56209b"), "Amount" : 100 } { "_id" : ObjectId("5e5231e9f8647eb59e56209c"), "Amount" : 500 } { "_id" : ObjectId("5e5231ebf8647eb59e56209d"), "Amount" : 400 }Following is the query to calculate sum ... Read More
 
 
			
			475 Views
To update only one property, use $addToSet in MongoDB. Let us create a collection with documents −> db.demo336.insertOne({"Name":"Chris", "Score":[45, 67, 78]}); { "acknowledged" : true, "insertedId" : ObjectId("5e522cb1f8647eb59e562097") } > db.demo336.insertOne({"Name":"David", "Score":[89, 93, 47]}); { "acknowledged" : true, "insertedId" : ObjectId("5e522cb2f8647eb59e562098") }Display all documents from a collection with the help of find() method −> db.demo336.find();This will produce the following output −{ "_id" : ObjectId("5e522cb1f8647eb59e562097"), "Name" : "Chris", "Score" : [ 45, 67, 78 ] } { "_id" : ObjectId("5e522cb2f8647eb59e562098"), "Name" : "David", "Score" : [ 89, 93, 47 ] }Following is the query to update only ... Read More