Duplicate a document in MongoDB using a new _id

Ok, I suppose that this is a silly question and probably has a simple answer.

How can I duplicate a document in MongoDB, changing the _id of the new one?

Imaging that you have the original document:

> var orig = db.MyCollection.findOne({_id: 'hi'})

And now I want another document in the collection with _id 'bye'.

Just change the id and re-insert.

> db.coll.insert({_id: 'hi', val: 1})
> var orig = db.coll.findOne({_id: 'hi'})
> orig._id = 'bye'
bye
> db.coll.insert(orig)
> db.coll.find()
{ "_id" : "hi", "val" : 1 }
{ "_id" : "bye", "val" : 1 }

You can give a new ObjectId to the copy Document. In mongo shell

var copy = db.collection.findOne();
for (var i = 0; i< 30; i++){
copy._id = new ObjectId();
db.collection.insert(copy);
}

A little improvement to the @689 response

var copy = db.collection.findOne({},{_id:0});
for (var i = 0; i< 30; i++){
db.collection.insert(copy);
}

You can use below code :

**Run using single line :**

db.collectionName.find({"_id" : ObjectId("5a4e47e0a21698d455000009")}).forEach(function(doc){var newDoc = doc; delete newDoc._id; db.collectionName.insert(newDoc); })

**a structural format for understanding :**

db.collectionName.find({"_id" : ObjectId("5a4e47e0a21698d455000009")}).forEach(function(doc){
var newDoc = doc;
delete newDoc._id;
db.collectionName.insert(newDoc);
})

In mongo shell: **It's OK**

db.products.find().forEach( function(doc){db.products.insert(
{
"price":doc.price,
"name":doc.name,
"typeName":doc.typeName,
"image":doc.image
}
)} );

single line
<!-- begin snippet: js hide: false console: true babel: false -->

<!-- language: lang-js -->

(function(doc){delete doc._id; db.collection.insert(doc)})(db.collection.findOne({ _id: ObjectId("60c9a684c7d51907c35ad463")}))

<!-- end snippet -->

structured
```
(function(doc) {
delete doc._id;
db.collection.insert(doc);
})(db.collection.findOne({ _id: ObjectId('60c9a684c7d51907c35ad463') }));

// (function(doc){...})(doc)
```

Many of the previous answers are simple and correct, but they are not efficient if your goal is to copy **all the documents in the collection** with new `_id`s. Here is a three-line solution that was more than 10x faster for me:

First copy the collection efficiently:
```
db.collection.aggregate([{ $out: "collection_copy" }]);
```

Now pass documents from the copy back into the original, but remove the `_id` field so a new `_id` is generated:
```
db.collection_copy.aggregate([{$project: {_id: 0}}, { $merge: "collection" }]);
```

Finally, clean up the copy:
```
db.collection_copy.drop();
```

With a pipeline

db.collection.insertMany(
db.collection.aggregate([
{
$match: { _id: ObjectId("") }
},

{ $unset: [ "_id"] }]).toArray());