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

Posted by: admin January 29, 2018 Leave a comment

Questions:
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?

Answers:

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.

Questions:
Answers:

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.


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.

Questions:
Answers:

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.

Questions:
Answers:

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)

Questions:
Answers:

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)

Questions:
Answers:

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.

Questions:
Answers:

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()

Questions:
Answers:

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.

Questions:
Answers:

Try this

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