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Get the top most document from a MongoDB collection
To get the topmost document, use find() along with limit(). To fetch only a single document, consider using limit(1). Let us create a collection with documents −
> db.demo681.insertOne({_id:101,Name:"Chris"});
{ "acknowledged" : true, "insertedId" : 101 }
> db.demo681.insertOne({_id:102,Name:"Bob"});
{ "acknowledged" : true, "insertedId" : 102 }
> db.demo681.insertOne({_id:103,Name:"David"});
{ "acknowledged" : true, "insertedId" : 103 }
> db.demo681.insertOne({_id:104,Name:"Bob"});
{ "acknowledged" : true, "insertedId" : 104 }
> db.demo681.insertOne({_id:105,Name:"Sam"});
{ "acknowledged" : true, "insertedId" : 105 }Display all documents from a collection with the help of find() method −
> db.demo681.find();
This will produce the following output −
{ "_id" : 101, "Name" : "Chris" }
{ "_id" : 102, "Name" : "Bob" }
{ "_id" : 103, "Name" : "David" }
{ "_id" : 104, "Name" : "Bob" }
{ "_id" : 105, "Name" : "Sam" }Following is the query to get the topmost element −
> db.demo681.find().limit(1);
This will produce the following output −
{ "_id" : 101, "Name" : "Chris" }
Updated on: 13-May-2020
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