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Django models default function run for every current object

models.py
class Subscription(models.Model):
#... many fields ...
# I added this field when I already had many objects
uniqueSubscriptionId = models.CharField(default=generateUniqueSubscription, max_length=30)
generateUniqueSubscription
from django.utils.crypto import get_random_string
def generateUniqueSubscription():
return get_random_string(20)
The Problem is that, when I run migrations, all of my old objects get the same uniqueSubscriptionId. I want each and every single old object to get a unique uniqueSubscriptionId.
How can I do that?

Here's what I did:
models.py
def updateOldSubscriptionObjs(apps, schema_editor):
old_subscription_model = apps.get_model("app_label", "Profile")
for obj in old_subscription_model.objects.all():
obj.uniqueSubscriptionId = generateUniqueSubscription()
obj.save()
class Subscription(models.Model):
#... many fields ...
# I added this field when I already had many objects
uniqueSubscriptionId = models.CharField(default=generateUniqueSubscription, max_length=30)
Then I ran makemigrations:
python manage.py makemigrations
Then edited the latest migration file:
class Migration(migrations.Migration):
dependencies = [
# forget this
]
operations = [
# .... many things ...
migrations.RunPython(updateOldProfileObjs)
]
Then ran migrate:
python manage.py migrate
And voila, all old objects got updated, and also, any new object will also get updated as I specified default.
If you are lazy like me, and don't want to do these things, then open django python shell:
python manage.py shell
and then execute this function in shell:
def updateOldSubscriptionObjs():
for obj in Subscription.objects.all():
obj.uniqueSubscriptionId = generateUniqueSubscription()
obj.save()
I wish if there was some built-in django feature for this.

ValueError: Cannot assign error in Django foreign key

I have three different Django apps in my project: questions, answers, and students. I created demo data for both Question and Student, and I want their objects to be foreign keys of Answer.
Below is the migration file of Answer.
from django.db import migrations
from django.apps import apps
questions = apps.get_model('questions', 'Question')
students = apps.get_model('students', 'Student')
def create_data(apps, schema_editor):
Answer = apps.get_model('answers', 'Answer')
Answer(question=questions.objects.get(pk="1"),
student=students.objects.get(pk="2"), answer="Main function").save()
class Migration(migrations.Migration):
dependencies = [
('answers', '0001_initial'),
]
operations = [
migrations.RunPython(create_data)
]
When I run python manage.py migrate, I get the error below.
ValueError: Cannot assign "<Question: What is this?>": "Answer.question" must be a "Question" instance.
Can please someone help me with this?

Check if there are questions and student rows with the ids that you need and then set the foreign keys using ids. Example:
answer = Answer(answer="Main function")
answer.question_id = 1
answer.student_id = 2
answer.save()
P.S. no need to import apps from django.apps, it's the first argument of create_data function. I suggest to use that instead.

Django model serialization problem with default fields

Inside of my app model, I use IntegerRangeField fields:
from django.db import models
from django.contrib.postgres.fields import IntegerRangeField
from django.contrib.postgres.validators import RangeMinValueValidator, RangeMaxValueValidator
from psycopg2.extras import NumericRange
class MyModel(models.Model):
...
field = IntegerRangeField(default=NumericRange(400, 600), validators=[
RangeMinValueValidator(1),
RangeMaxValueValidator(1000)
])
...
The "default" attributes are used in the admin panel UI only, and are not needed anywhere else.
If I add them after migration, they work smoothly. However, if I add them before I run makemigrations, I get this message:
ValueError: Cannot serialize: NumericRange(400, 600, '[)') There are
some values Django cannot serialize into migration files.
I don't even want the default values to be saved to my PostgreSQL database, I just want to not have to remove and bring them back every time I run makemigrations.
Any ideas?
(Didn't work: a custom object with "lower" and "higher" attributes, a single integer, a string, a tuple)
Python: 3.6.6, Django: 2.1.2, PostgreSQL: 11.0

Try to move default value calculation into separate function:
def get_default_range():
return NumericRange(400, 600)
class MyModel(models.Model):
field = IntegerRangeField(default=get_default_range, validators=[
RangeMinValueValidator(1),
RangeMaxValueValidator(1000)
])
In this case migration was successfully generated:
operations = [
migrations.AddField(
model_name='comment',
name='field',
field=django.contrib.postgres.fields.ranges.IntegerRangeField(
default=play.models.get_default_range,
validators=[django.contrib.postgres.validators.RangeMinValueValidator(1),
django.contrib.postgres.validators.RangeMaxValueValidator(1000)]),
),
]

I was able to solve this problem using the string representation of the range:
IntegerRangeField(default='[400, 600]')
django==3.0.5
psycopg2==2.8.5

EDIT I should point out that the original question was 2 years old, but at least in django 3.1, their is a serializer that you must register separately.
You need to register the serializer that is provided by django.
from psycopg2.extras import NumericRange
from django.contrib.postgres.serializers import RangeSerializer
from django.db.migrations.writer import MigrationWriter
MigrationWriter.register_serializer(NumericRange, RangeSerializer)
This piece was not in the documentation, but then you can add your defaults as you'd expect:
class AgeDivision(models.Model):
name = models.CharField(max_length=50, unique=True)
age_range = fields.IntegerRangeField(
unique=True, blank=True, default=NumericRange(None, None))
as for where to put this, it just needs to go along side any module that is only loaded once. The documentation didn't specify where to put custom serializers (as least that I could find), but I'd say put them in the migrations/__init__.py file for any app that requires the serializer. here's the documentation on migration serialization: https://docs.djangoproject.com/en/3.1/topics/migrations/#custom-serializers

Django Custom Migration Not Executing

So I added a new "status" field to a django database table. This field needed a default value, so I defaulted it to "New", but I then added a custom migration file that calls the save() method on all of the objects in that table, as I have the save() overridden to check a different table and pull the correct status from that. However, after running this migration, all of the statuses are still set to "New", so it looks like the save isn't getting executed. I tested this by manually calling the save on all the objects after running the migration, and the statuses are updated as expected.
Here's the table model in models.py:
class SOS(models.Model):
number = models.CharField(max_length=20, unique=True)
...
# the default="New" portion is missing here because I have a migration to remove it after the custom migration (shown below) that saves the models
status = models.CharField(max_length=20)
def save(self, *args, **kwargs):
self.status = self.history_set.get(version=self.latest_version).status if self.history_set.count() != 0 else "New"
super(SOS, self).save(*args, **kwargs)
And here is the migration:
# Generated by Django 2.0.5 on 2018-05-23 13:50
from django.db import migrations, models
def set_status(apps, schema_editor):
SOS = apps.get_model('sos', 'SOS')
for sos in SOS.objects.all():
sos.save()
class Migration(migrations.Migration):
dependencies = [
('sos', '0033_auto_20180523_0950'),
]
operations = [
migrations.RunPython(set_status),
]
So it seems pretty clear to me that I'm doing something wrong with the migration, but I matched it exactly to what I see in the Django Documentation and I also compared it to this StackOverflow answer, and I can't see what I'm doing wrong. There are no errors when I run the migrations, but the custom one I wrote does run pretty much instanteously, which seems strange, as when I do the save manually, it takes about 5 seconds to save all 300+ entries.
Any suggestions?
P.S. Please let me know if there are any relevant details I neglected to include.

When you run migrations and get Model from apps you can not use custom managers or custom save or create or something like that. This model only have the fields and that's all. If you want to achieve what you want you should add your logic into you migrations like this:
# comment to be more than 6 chars...
def set_status(apps, schema_editor):
SOS = apps.get_model('sos', 'SOS')
for sos in SOS.objects.all():
if sos.history_set.exists():
sos.status = sos.history_set.get(version=sos.latest_version).status
else:
sos.status = "New"
sos.save()

Django migration with uuid field generates duplicated values

I have a uuid field (not a primary key). The generated migration is:
from __future__ import unicode_literals
from django.db import migrations, models
import uuid
class Migration(migrations.Migration):
dependencies = [
....
]
operations = [
...
migrations.AddField(
model_name='device',
name='uuid',
field=models.UUIDField(default=uuid.uuid4, unique=True),
),
...
]
But when doing python manage.py migrate it is crashing with:
django.db.utils.IntegrityError: could not create unique index
"restaurants_device_uuid_key" DETAIL: Key
(uuid)=(f3858ded-b8e0-4ac0-8436-8a61b10efc73) is duplicated.
Strangely enough, the problem does not seem to occur with primary keys (which are maybe created by the database, and not internally by django?)
How can I add a uuid field, and make sure that migrations work?

Here is an example doing everything in one single migration thanks to a RunPython call.
# -*- coding: utf-8 -*
from __future__ import unicode_literals
from django.db import migrations, models
import uuid
def create_uuid(apps, schema_editor):
Device = apps.get_model('device_app', 'Device')
for device in Device.objects.all():
device.uuid = uuid.uuid4()
device.save()
class Migration(migrations.Migration):
dependencies = [
('device_app', 'XXXX'),
]
operations = [
migrations.AddField(
model_name='device',
name='uuid',
field=models.UUIDField(blank=True, null=True),
),
migrations.RunPython(create_uuid),
migrations.AlterField(
model_name='device',
name='uuid',
field=models.UUIDField(unique=True)
)
]

(Answer taken from the first comment)
See the django docs - Migrations that add unique fields
They recommend changing your single migration into three separate migrations:
Create field, set to null but not unique
Generate unique UUIDs
Alter the field to be unique

In the mode, you have configured, that you want unique values for the uuid fields, but with default values(the same for all). So if you have two 'device' objects in the database, the migrations add 'uuid' field to them with the default 'uuid.uuid4' value and when it tries to set it to the second one, it crashes because of the unique constrains.
If you drop your db and create new objects probably there will be not problems but thats not a solution for production db obviously :D.
A better solution is to create a data migration which sets different uuid value (generated by the default 'uuid' library) to every existing object in the database. You can read more about data migrations here:
https://docs.djangoproject.com/en/1.10/topics/migrations/#data-migrations
Then, when you create new objects, django will generate different uuid automatically. ;)
For the primary keys: Django adds it to the model by default.

You can provide a management command to populate the uuid field after uu_id column is created in the model but this has to be done after migrating the model and setting the field default as None:
from django.core.management.base import BaseCommand
from django.apps import apps
import uuid
class Command(BaseCommand):
def handle(self, *args, **options):
classes()
def classes():
app_models = apps.get_app_config('appname').get_models()
for model in app_models:
field = None
try:
field = model._meta.get_field('uu_id')
except:
pass
if field:
uu_id_list = list(model.objects.all().values_list('uu_id',flat=True))
if None in uu_id_list:
for row in model.objects.all():
row.uu_id = uuid.uuid4()
row.save()