How do I clone a Django model instance object and save it to the database?

Question

Foo.objects.get(pk="foo")
<Foo: test>

In the database, I want to add another object which is a copy of the object above.

Suppose my table has one row. I want to insert the first row object into another row with a different primary key. How can I do that?

Answer 1

Just change the primary key of your object and run save().

obj = Foo.objects.get(pk=<some_existing_pk>)
obj.pk = None
obj.save()

If you want auto-generated key, set the new key to None.

More on UPDATE/INSERT here.

Official docs on copying model instances: https://docs.djangoproject.com/en/2.2/topics/db/queries/#copying-model-instances

Answer 2

The Django documentation for database queries includes a section on copying model instances. Assuming your primary keys are autogenerated, you get the object you want to copy, set the primary key to None, and save the object again:

blog = Blog(name='My blog', tagline='Blogging is easy')
blog.save() # blog.pk == 1

blog.pk = None
blog.save() # blog.pk == 2

In this snippet, the first save() creates the original object, and the second save() creates the copy.

If you keep reading the documentation, there are also examples on how to handle two more complex cases: (1) copying an object which is an instance of a model subclass, and (2) also copying related objects, including objects in many-to-many relations.

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Note on miah's answer: Setting the pk to None is mentioned in miah's answer, although it's not presented front and center. So my answer mainly serves to emphasize that method as the Django-recommended way to do it.

Historical note: This wasn't explained in the Django docs until version 1.4. It has been possible since before 1.4, though.

Possible future functionality: The aforementioned docs change was made in this ticket. On the ticket's comment thread, there was also some discussion on adding a built-in copy function for model classes, but as far as I know they decided not to tackle that problem yet. So this "manual" way of copying will probably have to do for now.

Answer 3

Be careful here. This can be extremely expensive if you're in a loop of some kind and you're retrieving objects one by one. If you don't want the call to the database, just do:

from copy import deepcopy

new_instance = deepcopy(object_you_want_copied)
new_instance.id = None
new_instance.save()

It does the same thing as some of these other answers, but it doesn't make the database call to retrieve an object. This is also useful if you want to make a copy of an object that doesn't exist yet in the database.

Answer 4

Use the below code :

from django.forms import model_to_dict

instance = Some.objects.get(slug='something')

kwargs = model_to_dict(instance, exclude=['id'])
new_instance = Some.objects.create(**kwargs)

Answer 5

There's a clone snippet here, which you can add to your model which does this:

def clone(self):
  new_kwargs = dict([(fld.name, getattr(old, fld.name)) for fld in old._meta.fields if fld.name != old._meta.pk]);
  return self.__class__.objects.create(**new_kwargs)

Answer 6

How to do this was added to the official Django docs in Django1.4

https://docs.djangoproject.com/en/1.10/topics/db/queries/#copying-model-instances

The official answer is similar to miah's answer, but the docs point out some difficulties with inheritance and related objects, so you should probably make sure you read the docs.

Answer 7

I've run into a couple gotchas with the accepted answer. Here is my solution.

import copy

def clone(instance):
    cloned = copy.copy(instance) # don't alter original instance
    cloned.pk = None
    try:
        delattr(cloned, '_prefetched_objects_cache')
    except AttributeError:
        pass
    return cloned

Note: this uses solutions that aren't officially sanctioned in the Django docs, and they may cease to work in future versions. I tested this in 1.9.13.

The first improvement is that it allows you to continue using the original instance, by using copy.copy. Even if you don't intend to reuse the instance, it can be safer to do this step if the instance you're cloning was passed as an argument to a function. If not, the caller will unexpectedly have a different instance when the function returns.

copy.copy seems to produce a shallow copy of a Django model instance in the desired way. This is one of the things I did not find documented, but it works by pickling and unpickling, so it's probably well-supported.

Secondly, the approved answer will leave any prefetched results attached to the new instance. Those results shouldn't be associated with the new instance, unless you explicitly copy the to-many relationships. If you traverse the the prefetched relationships, you will get results that don't match the database. Breaking working code when you add a prefetch can be a nasty surprise.

Deleting _prefetched_objects_cache is a quick-and-dirty way to strip away all prefetches. Subsequent to-many accesses work as if there never was a prefetch. Using an undocumented property that begins with an underscore is probably asking for compatibility trouble, but it works for now.

Answer 8

This is yet another way of cloning the model instance:

d = Foo.objects.filter(pk=1).values().first()   
d.update({'id': None})
duplicate = Foo.objects.create(**d)

Answer 9

setting pk to None is better, sinse Django can correctly create a pk for you

object_copy = MyObject.objects.get(pk=...)
object_copy.pk = None
object_copy.save()

Answer 10

This does an in-memory copy that you can mutate independently.

original = CheckoutItem(title="test", ...)
copy = CheckoutItem()

for f in CheckoutItem._meta.fields:
   setattr(copy, f.attname, getattr(original, f.attname))

Or, as a method:

    def clone(self):
        """Returns a clone of this instance."""

        clone = self.__class__()
        for f in self.__class__._meta.fields:
            setattr(clone, f.attname, getattr(self, f.attname))

        return clone

Answer 11

To clone a model with multiple inheritance levels, i.e. >= 2, or ModelC below

class ModelA(models.Model):
    info1 = models.CharField(max_length=64)

class ModelB(ModelA):
    info2 = models.CharField(max_length=64)

class ModelC(ModelB):
    info3 = models.CharField(max_length=64)

Please refer the question here.

Answer 12

Try this

original_object = Foo.objects.get(pk="foo")
v = vars(original_object)
v.pop("pk")
new_object = Foo(**v)
new_object.save()

Answer 13

There is a package that can do this which creates a UI within the django admin site: https://github.com/RealGeeks/django-modelclone

pip install django-modelclone

Add "modelclone" to INSTALLED_APPS and import it within admin.py.

Then, whenever you want to make a model clonable, you just replace "admin.ModelAdmin" in the given admin model class "modelclone.ClonableModelAdmin". This results in a "Duplicate" button appearing within the instance details page for that given model.

Answer 14

If you have a OneToOneField then you should do it this way:

    tmp = Foo.objects.get(pk=1)
    tmp.pk = None
    tmp.id = None
    instance = tmp

Answer 15

This simple process works fine for me:

foo_obj = Foo.objects.get(pk="foo")
foo_values = foo_obj.__dict__
foo_values.pop('_state')
foo_values.pop('id')
foo_new_obj = Foo(**foo_values)
foo_new_obj.save()