You can use $set operator for this Let us first create a collection with documents −
> db.updateRecordDemo.insertOne({"StudentName":"Larry"}); { "acknowledged" : true, "insertedId" : ObjectId("5cbd6f95de8cc557214c0e0a") } > db.updateRecordDemo.insertOne({"StudentName":"David"}); { "acknowledged" : true, "insertedId" : ObjectId("5cbd6f9cde8cc557214c0e0b") } > db.updateRecordDemo.insertOne({"StudentName":"Mike"}); { "acknowledged" : true, "insertedId" : ObjectId("5cbd6f9dde8cc557214c0e0c") }
Display all documents from a collection with the help of find() method −
> db.updateRecordDemo.find().pretty();
This will produce the following output −
{ "_id" : ObjectId("5cbd6f95de8cc557214c0e0a"), "StudentName" : "Larry" } { "_id" : ObjectId("5cbd6f9cde8cc557214c0e0b"), "StudentName" : "David" } { "_id" : ObjectId("5cbd6f9dde8cc557214c0e0c"), "StudentName" : "Mike" }
Following is the query to update record in MongoDB without replacing the existing fields −
> db.updateRecordDemo.update({"_id" :ObjectId("5cbd6f9cde8cc557214c0e0b") },{$set : {"StudentAge":24}}); WriteResult({ "nMatched" : 1, "nUpserted" : 0, "nModified" : 1 })
Let us Display all documents from the collection once again −
> db.updateRecordDemo.find().pretty();
This will produce the following output −
{ "_id" : ObjectId("5cbd6f95de8cc557214c0e0a"), "StudentName" : "Larry" } { "_id" : ObjectId("5cbd6f9cde8cc557214c0e0b"), "StudentName" : "David", "StudentAge" : 24 } { "_id" : ObjectId("5cbd6f9dde8cc557214c0e0c"), "StudentName" : "Mike" }