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" }