Can UUIDField be used as default_auto_field in Django 3.2.^? - python

I have been using UUIDField as my primary key in Django. My project has a hierarchy of models with inherited fields, and at the top, I have the following superclass:
import uuid
from django.db import models
class HasIDMixin(models.Model):
id = models.UUIDField(primary_key=True, default=uuid.uuid4, editable=False, unique=True, name='id')
After updating to Django 3.2.4, I get the following warning:
WARNINGS:
app_base.MyProject: (models.W042) Auto-created primary key used when not defining a primary key type, by default 'django.db.models.AutoField'.
HINT: Configure the DEFAULT_AUTO_FIELD setting or the AppBaseConfig.default_auto_field attribute to point to a subclass of AutoField, e.g. 'django.db.models.BigAutoField'.
Following the advice of the warning, I tried both the DEFAULT_AUTO_FIELD in settings.py and the default_auto_field in the app_config and I get the following error:
ValueError: Primary key 'django.db.models.UUIDField' referred by DEFAULT_AUTO_FIELD must subclass AutoField.
I have seen others approach this problem with a custom child class to both UUIDField and AutoField (https://code.djangoproject.com/ticket/32577) but no working solution has been posted. Is this currently possible in Django 3.2.^? If not should I find a different primary key solution or roll back?

When one does not set a model field to be the primary key for that model Django will automatically add a field to your model named id which will be used as the primary key. This automatically added key is usually AutoField. From Django 3.2 onwards Django allows you to set what type of AutoField will be used for these automatically generated primary key fields by setting DEFAULT_AUTO_FIELD.
Considering you explicitly set a primary key for your model Django will not generate a field for it. But still you should specify DEFAULT_AUTO_FIELD as you might have some model where you don't specify the primary key. You can safely keep your UUIDField in your model as it is and also set DEFAULT_AUTO_FIELD like so:
DEFAULT_AUTO_FIELD = 'django.db.models.AutoField' # Or one of the other options as per your choice

Related

Reverse accessor clasher with Django [duplicate]

D:\zjm_code\basic_project>python manage.py syncdb
Error: One or more models did not validate:
topics.topic: Accessor for field 'content_type' clashes with related field 'Cont
entType.topic_set'. Add a related_name argument to the definition for 'content_t
ype'.
topics.topic: Accessor for field 'creator' clashes with related field 'User.crea
ted_topics'. Add a related_name argument to the definition for 'creator'.
topics.topic: Reverse query name for field 'creator' clashes with related field
'User.created_topics'. Add a related_name argument to the definition for 'creato
r'.
topicsMap.topic: Accessor for field 'content_type' clashes with related field 'C
ontentType.topic_set'. Add a related_name argument to the definition for 'conten
t_type'.
topicsMap.topic: Accessor for field 'creator' clashes with related field 'User.c
reated_topics'. Add a related_name argument to the definition for 'creator'.
topicsMap.topic: Reverse query name for field 'creator' clashes with related fie
ld 'User.created_topics'. Add a related_name argument to the definition for 'cre
ator'.
You have a number of foreign keys which django is unable to generate unique names for.
You can help out by adding "related_name" arguments to the foreignkey field definitions in your models.
Eg:
content_type = ForeignKey(Topic, related_name='topic_content_type')
See here for more.
http://docs.djangoproject.com/en/dev/ref/models/fields/#django.db.models.ForeignKey.related_name
Example:
class Article(models.Model):
author = models.ForeignKey('accounts.User')
editor = models.ForeignKey('accounts.User')
This will cause the error, because Django tries to automatically create a backwards relation for instances of accounts.User for each foreign key relation to user like user.article_set. This default method is ambiguous. Would user.article_set.all() refer to the user's articles related by the author field, or by the editor field?
Solution:
class Article(models.Model):
author = models.ForeignKey('accounts.User', related_name='author_article_set')
editor = models.ForeignKey('accounts.User', related_name='editor_article_set')
Now, for an instance of user user, there are two different manager methods:
user.author_article_set — user.author_article_set.all() will return a Queryset of all Article objects that have author == user
user.editor_article_set — user.editor_article_set.all() will return a Queryset of all Article objects that have editor == user
Note:
This is an old example — on_delete is now another required argument to models.ForeignKey. Details at What does on_delete do on Django models?
"If a model has a ForeignKey, instances of the foreign-key model will have access to a Manager that returns all instances of the first model. By default, this Manager is named FOO_set, where FOO is the source model name, lowercased."
But if you have more than one foreign key in a model, django is unable to generate unique names for foreign-key manager.
You can help out by adding "related_name" arguments to the foreignkey field definitions in your models.
See here:
https://docs.djangoproject.com/en/dev/topics/db/queries/#following-relationships-backward
If your models are inheriting from the same parent model, you should set a unique related_name in the parent's ForeignKey. For example:
author = models.ForeignKey('accounts.User', related_name='%(app_label)s_%(class)s_related')
It's better explained in th
If your models are inheriting from the same parent model, you should set a unique related_name in the parent's ForeignKey. For example:
author = models.ForeignKey('accounts.User', related_name='%(app_label)s_%(class)s_related')
It's better explained in the docs
I had a similar problem when I was trying to code a solution for a table that would pull names of football teams from the same table.
My table looked like this:
hometeamID = models.ForeignKey(Team, null=False, on_delete=models.CASCADE)
awayteamID = models.ForeignKey(Team, null=False, on_delete=models.CASCADE)
making the below changes solved my issue:
hometeamID = models.ForeignKey(Team, null=False, on_delete=models.CASCADE,related_name='home_team')
awayteamID = models.ForeignKey(Team, null=False, on_delete=models.CASCADE,related_name='away_team')
But in my case i am create a separate app for some functionality with same model name and field ( copy/paste ;) ) that's because of this type of error occurs i am just deleted the old model and code will work fine
May be help full for beginners like me :)
This isn't an ultimate answer for the question, however for someone it may solve the problem.
I got the same error in my project after checking out a really old commit (going to detached head state) and then getting the code base back up to date. Solution was to delete all *.pyc files in the project.
Do as the error message instructs you to:
Add a related_name argument to the
definition for 'creator'.

How to make a username as a foreign key of another model in django

I am very new to Django. I want to link a model which has 2 field 'username' and 'password'. I want to make 'username' field as as Foreign in another model. But as per Django we can only pass the whole Model Object, who is referring to as it's foreign key.
am I wrong somewhere? please give me any solution regarding this basic problem.
No you can link to any unique field of a Django model. So if your models look like:
class Target(models.Model):
name = models.CharField(max_length=128, unique=True)
class SourceModel(models.Model):
target = models.ForeignKey(Target, to_field='name', on_delete=models.CASCADE)
You can assign the value of the target column to the target_id then. So for example:
Target.objects.create(name='target1')
SourceModel.objects.create(target_id='target1')
So you do not need to pass a Target object itself. You can use the …_id "twin" field to use the target column value. The database will normally enforce referential integrity, and thus will prevent passing a non-existing value to the foreign key column.

Django. Create composite primary key

I want create composite promary key like {id, project_id}. I remove old tables(all). when i do:
python manage.py makemigrations
I have a mistake:
AssertionError: Model mdm.Group can't have more than one AutoField.
change my model:
id = models.AutoField(db_index=True, primary_key=False)
and add composite primary key as
constraints = [
models.UniqueConstraint(
fields=['id', 'project_id'], name='unique_group_project'
)
]
From docs:
By default, Django gives each model the following field:
id = models.AutoField(primary_key=True)
This is an auto-incrementing primary key.
If you’d like to specify a custom primary key, just specify primary_key=True on one of your fields. If Django sees you’ve explicitly set Field.primary_key, it won’t add the automatic id column.
Each model requires exactly one field to have primary_key=True (either explicitly declared or automatically added).
I just don't understand the problem. If I add AutoField, It's necessarily must PK. How I can resolve problem with Autofield id and composite PK (id, project_id)?

Can I create model in Django without automatic ID?

I need a table without a primary key (in Django it was created automatically). So my question is: Can I create a model without an ID/primary key?
I'm using Django 1.7.
You can create a model without an auto-incrementing key, but you cannot create one without a primary key.
From the Django Documentation:
If Django sees you’ve explicitly set Field.primary_key, it won’t add the automatic id column.
Each model requires exactly one field to have primary_key=True (either explicitly declared or automatically added).
No, you can't. Excerpt from the documentation:
Each model requires exactly one field to have primary_key=True (either explicitly declared or automatically added).
I've found the solution.
Since django need Primary Key (either it's composite or single-field ID) so, I've tried to set primary_key=True in every fields in its composite-key combination, and add those fields in Meta and groups in unique_together
class ReportPvUv(models.Model):
report_id = models.ForeignKey(Reports, primary_key=True)
rdate = models.DateField(primary_key=True)
fdate = models.DateTimeField(primary_key=True)
ga_pv = models.BigIntegerField()
ga_uv = models.BigIntegerField()
ur_pv = models.BigIntegerField()
ur_uv = models.BigIntegerField()
da_pv = models.BigIntegerField()
da_uv = models.BigIntegerField()
class Meta:
db_table = 'report_pv_uv'
unique_together = ('report_id', 'rdate', 'fdate')
and when I run makemigrations, there are no ID field in it's migrations script :D
thanks everybody

how to update model in django by adding new primary key field?

How to replace default primary key in Django model with custom primary key field?
I have a model with no primary key defined at first since django automatically adds an id field by default as primary field.
#models.py
from django.db import models
class Event(models.Model):
title = models.CharField(max_length=50, unique=True)
description = models.CharField(max_length=150)
I added some objects into it from django shell.
>>e = Event('meeting', 'Contents about meeting')
>>e.save()
>>e = Event('party', 'Contents about party')
>>e.save()
Then I require to add custom character field as primary into this model.
class Event(models.Model):
event-id = models.CharField(max_length=50, primary_key=True)
...
Running makemigrations:
$ python manage.py makemigrations
You are trying to add a non-nullable field 'event-id' to event without a default; we can't do that (the database needs something to populate existing rows).
Please select a fix:
1) Provide a one-off default now (will be set on all existing rows)
2) Quit, and let me add a default in models.py
Select an option: 1
Please enter the default value now, as valid Python
The datetime and `django.utils.timezone modules` are available, so you can do e.g. timezone.now()
>>> 'meetings'
Migrations for 'blog':
0002_auto_20141201_0301.py:
- Remove field id from event
- Add field event-id to event
But while running migrate it threw an error:
.virtualenvs/env/local/lib/python2.7/site-packages/django/db/backends/sqlite3/base.py", line 485, in execute
return Database.Cursor.execute(self, query, params)
django.db.utils.IntegrityError: UNIQUE constraint failed: blog_event__new.event-id
In my experience (using Django 1.8.* here), I've seen similar situations when trying to update the PK field for models that already exist, have a Foreign Key relationship to another model, and have associated data in the back-end table.
You didn't specify if this model is being used in a FK relation, but it seems this is the case.
In this case, the error message you're getting is because the data that already exists needs to be made consistent with the changes you're requesting --i.e. a new field will be the PK. This implies that the current PK must be dropped for django to 'replace' them. (Django only supports a single PK field per model, as per docs[1].)
Providing a default value that matches currently existing data in the related table should work.
For example:
class Organization(models.Model):
# assume former PK field no longer here; name is the new PK
name = models.CharField(primary_key=True)
class Product(models.Model):
name = models.CharField(primary_key=True)
organization = models.ForeignKey(Organization)
If you're updating the Organization model and products already exist, then existing product rows must be updated to refer to a valid Organization PK value. During the migration, you'd want to choose one of the existing Organization PKs (e.g. 'R&D') to update the existing products.
[1] https://docs.djangoproject.com/en/1.8/topics/db/models/#automatic-primary-key-fields
Django has already established an auto incrementing integer id as primary key in your backend as and when u made the previous model.
When u were trying to run the new model , An attempt was made to recreate a new primary key column that failed.
Another reason is,When u made the field,Django was expecting a unique value be explicitly defined for each new row which it couldn't found,hence the reason.
As told in previous answer you can re-create the migration and then try doing it again.It should work.. cheers :-)
The problem is that you made the field unique, then attempted to use the same value for all the rows in the table. I'm not sure if there's a way to programmatically provide the key, but you could do the following:
Delete the migration
Remove the primary_key attribute from the field
Make a new migration
Apply it
Fill in the value for all your rows
Add the primary_key attribute to the field
Make a new migration
Apply it
It's bruteforce-ish, but should work well enough.
Best of luck!

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