How to set the default of a JSONField to empty list in Django and django-jsonfield?

Question

Question

What is the best way to set a JSONField to have the default value of a new list in django?

Context

There is a model where one of the fields is a list of items. In the case where there are no items set, the model should have an empty list.

Current solution

from django.models import Model

class MyModel(Model):
    the_list_field = JSONField(default=[])

Is this the best way to do it? Should it be switched to use list instead?

Thanks!

Tidbit

Is there any reason why you're not using Django's built-in jsonfield? – Headband

@RodXavier MySQL < 5.7 and other legacy code uses them, so for consistency. – Tidbit

Answer 1

According to the Django documentation for JSONField you should indeed use default=list because using default=[] would create a mutable object that is shared between all instances of your field and could lead to some objects not having an empty list as a default.

Please note that this does not only apply for django.contrib.postgres.fields.JSONField but for all other kinds of objects and functions in Python in general.

Quote from the docs:

If you give the field a default, ensure it’s a callable such as list (for an empty default) or a callable that returns a list (such as a function). Incorrectly using default=[] creates a mutable default that is shared between all instances of

Ringo

Answer 2

list and dict are callable, while [] and {} are not (you can't do []()). So:

• Use JSONField(default=list) over JSONField(default=[])
• Use JSONField(default=dict) over JSONField(default={})

If you want to instantiate with some data you can do the following:

def jsonfield_default_value():  # This is a callable
    return [0, 0]  # Any serializable Python obj, e.g. `["A", "B"]` or `{"price": 0}`

class MyModel(Model):
    the_list_field = JSONField(default=jsonfield_default_value)

Kohinoor

Answer 3

It is true that you should provide a callable, however you do not need to explicitly define a function outside of the context of the model. Instead, you can take advantage of Python's functools.partial method to produce the following:

import functools

class MyModel(Model):
    the_json_field = JSONField(default=partial(dict, (("a", "A"), ("b", "B")))

Basilica