How to sort documents and display only a single field in MongoDB 4?

To sort documents in MongoDB 4, use sort(). To display only a single sorted field, set this to 1.

Let us create a collection containing documents -

> db.demo611.insertOne({"Name":"Chris"});{
   "acknowledged" : true, "insertedId" : ObjectId("5e987110f6b89257f5584d83")
}
> db.demo611.insertOne({"Name":"Adam"});{
   "acknowledged" : true, "insertedId" : ObjectId("5e987115f6b89257f5584d84")
}
> db.demo611.insertOne({"Name":"John"});{
   "acknowledged" : true, "insertedId" : ObjectId("5e987118f6b89257f5584d85")
}
> db.demo611.insertOne({"Name":"Bob"});{
   "acknowledged" : true, "insertedId" : ObjectId("5e98711bf6b89257f5584d86")
}

Display all the documents in the collection with the help of find() method-

> db.demo611.find();
This will produce the following output:
{ "_id" : ObjectId("5e987110f6b89257f5584d83"), "Name" : "Chris" }
{ "_id" : ObjectId("5e987115f6b89257f5584d84"), "Name" : "Adam" }
{ "_id" : ObjectId("5e987118f6b89257f5584d85"), "Name" : "John" }
{ "_id" : ObjectId("5e98711bf6b89257f5584d86"), "Name" : "Bob" }

The following is how to sort the documents using MongoDB 4 The query -

> db.demo611.find().sort({Name:1});

will produce the following output&mius;

{ "_id" : ObjectId("5e987115f6b89257f5584d84"), "Name" : "Adam" }
{ "_id" : ObjectId("5e98711bf6b89257f5584d86"), "Name" : "Bob" }
{ "_id" : ObjectId("5e987110f6b89257f5584d83"), "Name" : "Chris" }
{ "_id" : ObjectId("5e987118f6b89257f5584d85"), "Name" : "John" }

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