How to return only a single property “_id” in MongoDB?

Following is the syntax to return only a single property _id in MongoDB

db.yourCollectionName.find({}, {"_id": 1}).pretty();

Let us first create a collection with documents

> db.singlePropertyIdDemo.insertOne({"_id":101,"UserName":"Larry","UserAge":21}); { "acknowledged" : true, "insertedId" : 101 } > db.singlePropertyIdDemo.insertOne({"_id":102,"UserName":"Mike","UserAge":26}); { "acknowledged" : true, "insertedId" : 102 } > db.singlePropertyIdDemo.insertOne({"_id":103,"UserName":"Chris","UserAge":24}); { "acknowledged" : true, "insertedId" : 103 } > db.singlePropertyIdDemo.insertOne({"_id":104,"UserName":"Robert","UserAge":23}); { "acknowledged" : true, "insertedId" : 104 } > db.singlePropertyIdDemo.insertOne({"_id":105,"UserName":"John","UserAge":27}); { "acknowledged" : true, "insertedId" : 105 }

Following is the query to display all documents from a collection with the help of find() method

> db.singlePropertyIdDemo.find().pretty();

This will produce the following output

{ "_id" : 101, "UserName" : "Larry", "UserAge" : 21 } { "_id" : 102, "UserName" : "Mike", "UserAge" : 26 } { "_id" : 103, "UserName" : "Chris", "UserAge" : 24 } { "_id" : 104, "UserName" : "Robert", "UserAge" : 23 } { "_id" : 105, "UserName" : "John", "UserAge" : 27 }

Following is the query to return only a single property _id

> db.singlePropertyIdDemo.find({}, {"_id": 1}).pretty();

This will produce the following output

{ "_id" : 101 } { "_id" : 102 } { "_id" : 103 } { "_id" : 104 } { "_id" : 105 }