I would like to change a name of specific fields in a model:
class Foo(models.Model):
name = models.CharField()
rel = models.ForeignKey(Bar)
should change to:
class Foo(models.Model):
full_name = models.CharField()
odd_relation = models.ForeignKey(Bar)
What's the easiest way to do this using South?
You can use the db.rename_column function.
class Migration:
def forwards(self, orm):
# Rename 'name' field to 'full_name'
db.rename_column('app_foo', 'name', 'full_name')
def backwards(self, orm):
# Rename 'full_name' field to 'name'
db.rename_column('app_foo', 'full_name', 'name')
The first argument of db.rename_column is the table name, so it's important to remember how Django creates table names:
Django automatically derives the name of the database table from the name of your model class and the app that contains it. A model's database table name is constructed by joining the model's "app label" -- the name you used in manage.py startapp -- to the model's class name, with an underscore between them.
In the case where you have a multi-worded, camel-cased model name, such as ProjectItem, the table name will be app_projectitem (i.e., an underscore will not be inserted between project and item even though they are camel-cased).
Here's what I do:
Make the column name change in your model (in this example it would be myapp/models.py)
Run ./manage.py schemamigration myapp renaming_column_x --auto
Note renaming_column_x can be anything you like, it's just a way of giving a descriptive name to the migration file.
This will generate you a file called myapp/migrations/000x_renaming_column_x.py which will delete your old column and add a new column.
Modify the code in this file to change the migration behaviour to a simple rename:
class Migration(SchemaMigration):
def forwards(self, orm):
# Renaming column 'mymodel.old_column_name' to 'mymodel.new_column_name'
db.rename_column(u'myapp_mymodel', 'old_column_name', 'new_column_name')
def backwards(self, orm):
# Renaming column 'mymodel.new_column_name' to 'mymodel.old_column_name'
db.rename_column(u'myapp_mymodel', 'new_column_name', 'old_column_name')
I didn't know about db.rename column, sounds handy, however in the past I have added the new column as one schemamigration, then created a datamigration to move values into the new field, then a second schemamigration to remove the old column
Django 1.7 introduced Migrations so now you don't even need to install extra package to manage your migrations.
To rename your model you need to create empty migration first:
$ manage.py makemigrations <app_name> --empty
Then you need to edit your migration's code like this:
from django.db import models, migrations
class Migration(migrations.Migration):
dependencies = [
('yourapp', 'XXXX_your_previous_migration'),
]
operations = [
migrations.RenameField(
model_name='Foo',
old_name='name',
new_name='full_name'
),
migrations.RenameField(
model_name='Foo',
old_name='rel',
new_name='odd_relation'
),
]
And after that you need to run:
$ manage.py migrate <app_name>
Just change the model and run makemigrations in 1.9
Django automatically detects that you've deleted and created a single field, and asks:
Did you rename model.old to model.new (a IntegerField)? [y/N]
Say yes, and the right migration gets created. Magic.
Add south to your installed apps in project setting file.
Comment out the added/modified field/table.
$ manage.py Schemamigration <app_name> --initial
$ manage.py migrate <app_name> --Fake
Un-comment the field and write the modified one
$ manage.py Schemamigration --auto
$ manage.py migrate <app_name>
If you are using 'pycharm', then you can use 'ctrl+shift+r' instead of 'manage.py' , and 'shift ' for parameters.
Related
I want to get the name of the last applied migration in Django. I know that Django migrations are stored in django_migrations table, however django.db.migrations.migration.Migration is not a models.Model backed by that table. This means you cannot do:
migration_info = Migration.objects.all()
Is there any built-in way of retrieveing the data from django_migrations, or should i just create my own read-only Model:
class MigrationInfo(models.Model):
class Meta:
managed = False
db_table = "django_migrations"
This works on Django 1.11/1.8/2.1 & 3.0.4:
from django.db.migrations.recorder import MigrationRecorder
last_migration = MigrationRecorder.Migration.objects.latest('id')
print(last_migration.app) # The app where the migration belongs
print(last_migration.name) # The name of the migration
There doesn't seem to be documentation for this command, but here you may find the source code which is documented properly.
To store information about applied migrations Django uses plain table and it is accessible as #classproperty through the MigrationRecorder class:
from django.db.migrations.recorder import MigrationRecorder
lm = MigrationRecorder.Migration.objects.filter(app='core').last()
It is also easy to retrieve this information from the command line:
Get the last applied migration for the particular app
python manage.py showmigrations --list <app_name> | grep "\[X\]" | tail -1
Get the ordered list of unapplied migrations
python manage.py showmigrations --plan | grep "\[ \]"
A lot easier, you could also parse out the last line of:
./manage.py showmigrations <app_name>
I have a Product model and I want to extend by using OneToOneField.
For example
class Product:
name = models.CharField(..)
price = models.FloatField(...)
I want to do like this
class MyProduct:
product = models.OneToOneField(myapp.Product, on_delete=models.CASCADE)
location = models.CharField(...)
and using signal
def create_myproduct(sender, instance, created, **kwargs):
"""Create MyProduct class for every new Product"""
if created:
MyProduct.objects.create(product=instance)
signals.post_save.connect(create_myproduct, sender=Product, weak=False,
dispatch_uid='models.create_myproduct')
This works for newly created Product, so I can do like this in template.
{{ product.myproduct.location }}
But Old products that created before adding this OneToOneRelation,has no field 'myproduct' and that template code didn't work.
I heard I need a data migrations for old product using RunPython or manage.py shell. Can you teach me how to do? I read a documentation from django, but still don't fully understand.
you can add new migration. and apply it.
something like this code:
# -*- coding: utf-8 -*-
# Generated by Django 1.11.2 on 2017-07-22 06:04
from __future__ import unicode_literals
from django.db import migrations, models
def create_myproducts(apps, schema_editor):
Product = apps.get_model('myapp', 'Product')
MyProduct = apps.get_model('myapp', 'MyProduct')
for prod in Product.objects.all():
MyProduct.objects.create(product=prod)
class Migration(migrations.Migration):
dependencies = [
('myapp', 'your last migration'),
]
operations = [
migrations.RunPython(create_myproducts)
]
I just found out.
Like Rohit Jain said
product.myproduct is None.
When I tried to access product.myproduct, I got an exception that object does not exist. It has a relation to myproduct but the actual object doesn't exist.
What I really want was creating MyProduct object and add it to Product class.
So I did it in python manage.py shell
products = Product.objects.all()
for prod in products:
if not hasattr(prod, 'myproduct'):
prod.myproduct = MyProduct.objects.create(product=prod)
prod.save()
I think it works for me now.
Thank you guys
You should just migrate your models in a normal way
python manage.py makemigrations
python manage.py migrate
During making migrations you will be asked how to fill new fields for existing data
Please notice that when you are using Django under 1.7 version you do not have migrations (and syncdb will not do the job for existing tables) - consider using the 3rd part tool like south
I'm doing a sqlite3 data migration using South. My old schema has the following model for UserProfile:
class UserProfile(models.Model):
user = models.OneToOneField(User)
weekOne = models.OneToOneField(WeekOne)
weekTwo = models.OneToOneField(WeekTwo)
weekThree = models.OneToOneField(WeekThree)
But I have sense added a number of new weeks, i.e., weekFour, weekFive, weekSix, etc. Every week is itself a model, inheriting from a generic Week model. So the basic prototype of a Week model looks like this:
class WeekOne(Week):
name = u'Week One'
exercises = ['squats', 'lunges', 'stairDaysCount', 'skipStairs']
# Required benchmarks for given exercises
squatBenchmark = 1000
lungeBenchmark = 250
stairDaysCountBenchmark = 3
totalGoals = 4
My question is, what kind of code should I put in my datamigration code so that I can populate old UserProfiles with the additional weeks. I had something like this in mind:
def forwards(self, orm):
for user in orm.UserProfile.objects.all():
user.weekFour = orm.WeekFour()
user.weekFive = orm.weekFive()
# etc.
But that doesn't seem to be working. I get this error when I try to run schema migration:
Migration 'my_app:0002_newWeeks' is marked for no-dry-run
And later this:
DatabaseError: no such column: my_app_userprofile.weekFour_id
Just run this in the terminal
python manage.py schemamigration myapp --auto
Then
python manage.py migrate
Following is my model:
class myUser_Group(models.Model):
name = models.CharField(max_length=100)
class Channel(models.Model):
name = models.CharField(max_length=100)
description = models.CharField(max_length=1000)
belongs_to_group = models.ManyToManyField(myUser_Group)
class Video(models.Model):
video_url = models.URLField(max_length=300)
belongs_to_channel = models.ManyToManyField(Channel)
description = models.CharField(max_length=1000)
tags = TagField()
class UserProfile(models.Model):
user = models.OneToOneField(User)
class User_History(models.Model):
date_time = models.DateTimeField()
user = models.ForeignKey(UserProfile, null=True, blank=True)
videos_watched = models.ManyToManyField(Video)
I just wanted to remove the underscores from all the class names so that User_History looks UserHistory, also the foreign keys should be updated. I tried using south but could not find it in the documentaion.
One way is export the data, uninstall south, delete migration, rename the table and then import data again. Is there any other way to do it?
You can do this using just South.
For this example I have an app called usergroups with the following model:
class myUser_Group(models.Model):
name = models.CharField(max_length=100)
which I assume is already under migration control with South.
Make the model name change:
class MyUserGroup(models.Model):
name = models.CharField(max_length=100)
and create an empty migration from south
$ python manage.py schemamigration usergroups model_name_change --empty
This will create a skeleton migration file for you to specify what happens. If we edit it so it looks like this (this file will be in the app_name/migrations/ folder -- usergroups/migrations/ in this case):
import datetime
from south.db import db
from south.v2 import SchemaMigration
from django.db import models
class Migration(SchemaMigration):
def forwards(self, orm):
# Change the table name from the old model name to the new model name
# ADD THIS LINE (using the correct table names)
db.rename_table('usergroups_myuser_group', 'usergroups_myusergroup')
def backwards(self, orm):
# Provide a way to do the migration backwards by renaming the other way
# ADD THIS LINE (using the correct table names)
db.rename_table('usergroups_myusergroup', 'usergroups_myuser_group')
models = {
'usergroups.myusergroup': {
'Meta': {'object_name': 'MyUserGroup'},
'id': ('django.db.models.fields.AutoField', [], {'primary_key': 'True'}),
'name': ('django.db.models.fields.CharField', [], {'max_length': '100'})
}
}
complete_apps = ['usergroups']
In the forwards method we are renaming the database table name to match what the django ORM will look for with the new model name. We reverse the change in backwards to ensure the migration can be stepped back if required.
Run the migration with no need to import/export the exisiting data:
$ python manage.py migrate
The only step remaining is to update the foreign key and many-to-many columns in the models that refer to myUser_Group and change to refer to MyUserGroup.
mmcnickle's solution may work and seems reasonable but I prefer a two step process. In the first step you change the table name.
In your model make sure you have your new table name in:
class Meta:
db_table = new_table_name'
Then like mmcnickle suggested, create a custom migration:
python manage.py schemamigration xyz migration_name --empty
You can read more about that here:
https://docs.djangoproject.com/en/dev/ref/models/options/
Now with your custom migration also add the line to rename your table forward and backwards:
db.rename_table("old_table_name","new_table_name")
This can be enough to migrate and change the table name but if you have been using the Class Meta custom table name before then you'll have to do a bit more. So I would say as a rule, just to be safe do a search in your migration file for "old_table_name" and change any entries you find to the new table name. For example, if you were previously using the Class Meta custom table name, you will likely see:
'Meta': {'object_name': 'ModelNameYouWillChangeNext', 'db_table': "u'old_table_name'"},
So you'll need to change that old table name to the new one.
Now you can migrate with:
python manage.py migrate xyz
At this point your app should run since all you have done is change the table name and tell Django to look for the new table name.
The second step is to change your model name. The difficulty of this really depends on your app but basically you just need to change all the code that references the old model name to code that references the new model name. You also probably need to change some file names and directory names if you have used your old model name in them for organization purposes.
After you do this your app should run fine. At this point your task is pretty much accomplished and your app should run fine with a new model name and new table name. The only problem you will run into using South is the next time you create a migration using it's auto detection feature it will try to drop the old table and create a new one from scratch because it has detected your new model name. To fix this you need to create another custom migration:
python manage.py schemamigration xyz tell_south_we_changed_the_model_name_for_old_model_name --empty
The nice thing is here you do nothing since you have already changed your model name so South picks this up. Just migrate with "pass" in the migrate forwards and backwards:
python manage.py migrate xyz
Nothing is done and South now realizes it is up to date. Try:
python manage.py schemamigration xyz --auto
and you should see it detects nothing has changed
Hey,
I've a database already created. Now I've updated UserProfile with:
class UserProfile(models.Model):
user = models.ForeignKey(User, unique = True, related_name = 'user')
follows = models.ManyToManyField("self", related_name = 'follows') <-- NEW LINE
so python manage.py sqlall myapp returns me:
[...]
CREATE TABLE "myapp_userprofile_follows" (
"id" integer NOT NULL PRIMARY KEY,
"from_userprofile_id" integer NOT NULL,
"to_userprofile_id" integer NOT NULL,
UNIQUE ("from_userprofile_id", "to_userprofile_id")
)
[...]
When I run python manage.py syncdb:
Creating tables ...
Installing custom SQL ...
Installing indexes ...
No fixtures found.
But the table is not created when I try to insert data into. Why? (I'm testing locally, with sqlite3)
manage.py syncdb will not modify existing tables to add or remove fields. You will need to either manually modify your database, or use a tool like South to create automated database migrations (which is what I highly recommend)
have you added your app to INSTALLED_APPS in settings.pys:
settings.py
INSTALLED_APPS = (
... ,
'my_app',
)
https://docs.djangoproject.com/en/dev/ref/settings/?from=olddocs#installed-apps