In django models we have option named managed which can be set True or False
According to documentation the only difference this option makes is whether table will be managed by django or not. Is management by django or by us makes any difference?
Is there any pros and cons of using one option rather than other?
I mean why would we opt for managed=False? Will it give some extra control or power which affects my code?
The main reason for using managed=False is if your model is backed by something like a database view, instead of a table - so you don't want Django to issue CREATE TABLE commands when you run syncdb.
Right from Django docs:
managed=False is useful if the model represents an existing table or a database view that has been created by some other means. This is the only difference when managed=False. All other aspects of model handling are exactly the same as normal
When ever we create the django model, the managed=True implicitly is
true by default. As we know that when we run python manage.py makemigrations the migration file(which we can say a db view) is
created in migration folder of the app and to apply that migration i.e
creates the table in db or we can say schema.
So by managed=False, we restrict Django to create table(scheme, update
the schema of the table) of that model or its fields specified in
migration file.
Why we use its?
case1: Sometime we use two db for the project for
example we have db1(default) and db2, so we don't want particular
model to be create the schema or table in db1 so we can this or we can
customize the db view.
case2. In django ORM, the db table is tied to django ORM model, It
help tie a database view to bind with a django ORM model.
Can also go through the link:
We can add our raw sql for db view in migration file.
The raw sql in migration look like: In 0001_initial.py
from future import unicode_literals
from django.db import migrations, models
class Migration(migrations.Migration):
initial = True
dependencies = [
]
operations = [
migrations.RunSQL(
CREATE OR REPLACE VIEW app_test AS
SELECT row_number() OVER () as id,
ci.user_id,
ci.company_id,
),
]
Above code is just for overview of the looking of the migration file, can go through above link for brief.
Related
If I have a PostgresDB that contains both Django models and other SQL tables, is it possible to register these other SQL tables in the Django admin panel?
More details about the setup:
I have a docker-compose setup where Django is running in one container, a Postgres DB in another, and a slack-app in a third container. Django is connected to the DB and the models are registered in the admin panel. This works as intended. The slack-app is also connected to the same DB and has some tables there that are not Django-models. I would like to also access these through the Django admin panel in order to have everything in one place. Is this possible?
You can define unmanaged models in Django. These models will not construct migrations, but will only query the database to select, insert, etc.
Django offers a tool inspectdb [Django-doc] to inspect the database and write the corresponding unamanged models. You thus can use this with:
python3 manage.py inspectdb table1 table2 tablen
It will then write the corresponding models for these tables to the standard output channel, and you thus can copy these in the models.py. In the Meta of these models it will add a managed = False to denote that Django will not migrate these models.
Once you registered these models, you can register a ModelAdmin with:
from django.contrib import admin
from app_name.models import Model1, Model2, Modeln
admin.site.register(Model1)
admin.site.register(Model2)
admin.site.register(Modeln)
I am migrating to django-oscar2.1
I have forked catalogue app, it has some existing migrations. Newly added field is_public is in migration #10. I have a custom migration #5 which is using create_from_breadcrumbs method. So when I tried to run migrate on fresh db migration#5 trigger error is_public field not exists in db.
create_from_breadcrumbs
Oscar loads a model with latest state which contains is_public in models but not in db.
In migration #5 I am loading model like this
Category = apps.get_model("catalogue", "Category")
You cannot use create_from_breadcrumbs in a migration like that. The reason for this is explained in Django's documentation on migrations - that the version of the model used in external code may be different from the one that this migration expects - which is why it fails.
If you want to create model objects as part of your migration you need to use the historic version of the model in the migration. This means defining your own version of create_from_breadcrumbs, and using the historic version of the model as suggested by the Django documentation:
def create_from_breadcrumbs_for_migration(apps, schema_editor):
Category = apps.get_model('catalogue', 'Category')
# Perform logic to create categories here.
class Migration(migrations.Migration):
dependencies = [
('catalogue', '0004_xxxx'),
]
operations = [
migrations.RunPython(create_from_breadcrumbs_for_migration),
]
That said, I would argue that migrations are not meant for this sort of data population in the first place. You would in my view be better off writing a separate management command for this, which you run after the migrations have been applied.
I added a field to one of my models, but in the 'models' folder I have two other python files which have only View models from which I query views in my database. When I run the makemigrations command, the new migrations file that is created includes also adding these view models to my database as tables (which I don't want to). How can I ignore these changes, and only commit the one addition of a field to an actual table on the database.
I think I maybe have to delete the migrations.CreateModel... in the new migrations file and only keep the migrations.addField... , then run the 'migrate' command. I didn't proceed with this because I'm not sure and maybe it will mess up my database in some way.
Thanks in advance to anyone who can help.
when you make a model for database view you must add meta class managed = false and db_table like this:
class MyViewModel(models.Model):
field: models.CharField(max_length=100)
class Meta:
managed = False
db_table = 'database_view_name'
when you write this and run makemigrations a migration generated contains this model but when you run migrate this migration doesnt change anything on database.
you also can create view using migrations in python. see migrations.RunPython for more details
Just a simple question.
After I connect my django app to a remote database, I don't need to use Model.py to create tables in the database, then what is the function for Model.py at that moment?
If you want to use the Django ORM, you'll need to create models in the models.py file that match your remote database. If you don't want django creating or deleting tables on this DB, the managed=False option needs to be set for each model.
https://docs.djangoproject.com/en/1.11/ref/models/options/#managed
As you said after running migrations all tables in models.py file will be created. Later on, if you want to do some database operations, you may be using Django ORM. If you don't have models.py you won't be able to do such operations.
For example:
To create an entry to the table MyModel.
from your_app.models import MyModel
MyModel.objects.create(<field_name>=<value>)
I hope this gives you some idea.
I use Django and South for my database. Now I want to add a new Model and a field in an existing model, referencing the new model. For example:
class NewModel(models.Model):
# a new model
# ...
class ExistingModel(models.Model):
# ... existing fields
new_field = models.ForeignKey(NewModel) # adding this now
Now South obviously complains that I added a non-null field and asks me to enter a one-off value. But what I really want is to create a new NewModel instance for every existing ExistingModel instance, thus fulfilling the database requirements. Is that possible somehow?
The easiest way to do this is to write a schema migration that makes the column change, and then write a datamigration to correctly fill in the value. Depending on the database you're using you'll have to do this in slightly different ways.
Sqlite
For Sqlite, you can add a sentinel value for the relation and use a datamigration to fill it in without any issue:
0001_schema_migration_add_foreign_key_to_new_model_from_existing_model.py
from south.db import db
from south.v2 import SchemaMigration
from django.db import models
class Migration(SchemaMigration):
def forwards(self, orm):
db.add_column('existing_model_table', 'new_model',
self.gf('django.db.models.fields.related.ForeignKey')(default=0, to=orm['appname.new_model']), keep_default=False)
0002_data_migration_for_new_model.py:
class Migration(DataMigration):
def forwards(self, orm):
for m in orm['appname.existing_model'].objects.all():
m.new_model = #custom criteria here
m.save()
This will work just fine, with no issues.
Postgres and MySQL
With MySql, you have to give it a valid default. If 0 isn't actually a valid Foreignkey, you'll get errors telling you so.
You could default to 1, but there are instances where that isn't a valid foreign key (happened to me because we have different environments, and some environments publish to other databases, so the IDs rarely match up (we use UUIDs for cross-database identification, as God intended).
The second issue you get is that South and MySQL don't play well together. Partially because MySQL doesn't have the concept of DDL transactions.
In order to get around some issues you will inevitably face (including the error I mentioned above and from South asking you to mark orm items in a SchemaMigration as no-dry-run), you need to change the above 0001 script to do the following:
0001_schema_migration_add_foreign_key_to_new_model_from_existing_model.py
from south.db import db
from south.v2 import SchemaMigration
from django.db import models
class Migration(SchemaMigration):
def forwards(self, orm):
id = 0
if not db.dry_run:
new_model = orm['appname.new_model'].objects.all()[0]
if new_model:
id = new_model.id
db.add_column('existing_model_table', 'new_model',
self.gf('django.db.models.fields.related.ForeignKey')(default=id, to=orm['appname.new_model']), keep_default=False)
And then you can run the 0002_data_migration_for_new_model.py file as normal.
I advise using the same example above for Postgres and for MySql. I don't remember any issues offhand with Postgres with the first example, but I'm certain the second example works for both (tested).
You want a data migration to supplement your schema migration in this scenario.
South has a nice step by step tutorial on how to achieve this in the docs, here.
It's not uncommon in South to have the desired outcome spread over two or three schema/data migrations as its not always possible to do it in one big hit (sometimes depends on the underlying db if it will tolerate adding a non null column with no default). So in this case you might add a schema migration that has a default, then a data migration with your object manipulation then a final schema migration.