How can I display only unique record from MongoDB and ignore the duplicates?

To display only unique records, use distinct() in MongoDB. Let us create a collection with documents −

> db.demo613.insertOne({"Name":"Chris"});{    "acknowledged" : true, "insertedId" : ObjectId("5e988bd4f6b89257f5584d88") } > db.demo613.insertOne({"Name":"Bob"});{    "acknowledged" : true, "insertedId" : ObjectId("5e988bdbf6b89257f5584d89") } > db.demo613.insertOne({"Name":"Bob"});{    "acknowledged" : true, "insertedId" : ObjectId("5e988bddf6b89257f5584d8a") } > db.demo613.insertOne({"Name":"David"});{    "acknowledged" : true, "insertedId" : ObjectId("5e988be0f6b89257f5584d8b") } > db.demo613.insertOne({"Name":"Chris"});{    "acknowledged" : true, "insertedId" : ObjectId("5e988be2f6b89257f5584d8c") } > db.demo613.insertOne({"Name":"Sam"});{    "acknowledged" : true, "insertedId" : ObjectId("5e988be8f6b89257f5584d8d") }

Display all documents from a collection with the help of find() method −

> db.demo613.find();

This will produce the following output −

{ "_id" : ObjectId("5e988bd4f6b89257f5584d88"), "Name" : "Chris" } { "_id" : ObjectId("5e988bdbf6b89257f5584d89"), "Name" : "Bob" } { "_id" : ObjectId("5e988bddf6b89257f5584d8a"), "Name" : "Bob" } { "_id" : ObjectId("5e988be0f6b89257f5584d8b"), "Name" : "David" } { "_id" : ObjectId("5e988be2f6b89257f5584d8c"), "Name" : "Chris" } { "_id" : ObjectId("5e988be8f6b89257f5584d8d"), "Name" : "Sam" }

Following is the query to get a unique record −

> db.demo613.distinct("Name");

This will produce the following output −

[ "Chris", "Bob", "David", "Sam" ]