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Get a single element from the array of results by index in MongoDB
To get a single element, use aggregation and LIMIT. The skip() is used to skip a specific number of documents.
Let us first create a collection with documents −
> db.demo391.insertOne(
... { "_id" : 101, "Name" : "Chris", Values: ["101", "102"] }
... )
{ "acknowledged" : true, "insertedId" : 101 }
> db.demo391.insertOne(
... { "_id" : 111, "Name" : "Chris", Values: ["101", "102"] }
... )
{ "acknowledged" : true, "insertedId" : 111 }
> db.demo391.insertOne(
... { "_id" : 121, "Name" : "Chris", Values: ["101", "102"] }
... )
{ "acknowledged" : true, "insertedId" : 121 }Display all documents from a collection with the help of find() method −
> db.demo391.find();
This will produce the following output −
{ "_id" : 101, "Name" : "Chris", "Values" : [ "101", "102" ] }
{ "_id" : 111, "Name" : "Chris", "Values" : [ "101", "102" ] }
{ "_id" : 121, "Name" : "Chris", "Values" : [ "101", "102" ] }Following is the query to get a single element from the array of results by index −
> var i=2;
> db.demo391.aggregate([
... { $match : {"Name": "Chris"}},
... { $skip : i-1},
... { $limit : 1 }
... ]);This will produce the following output −
{ "_id" : 111, "Name" : "Chris", "Values" : [ "101", "102" ] }
Updated on: 02-Apr-2020
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