I have a Django application with a My-SQL database. recently I alter the table_name with the help of MySQL query in the MySQL-shell, after this when I run makemigration and migrate command terminal says "No changes detected". how can i resolve this issue and create again this table with help of Django makemigration and migrate?
can I delete a table from MySQL, any possibility will Django create it again?
If you renamed your table outside Django - you will have to tell Django the new table name like so (using the Meta class):
class Model(models.Model):
name = models.CharField(max_length=255)
class Meta:
db_table = 'new_table_name'
To re-create your table using existing model you need to reset migration for that app to zero and then run migration again
python manage.py migrate APP_NAME zero
python manage.py migrate APP_NAME
It's because the migrations table managed by django doesn't reflect the correct db schema since it was already modified outside of django. If you don't have any important data you can do a migration rollback or recreate the table by hand.
The best way to dela with this is to rename your table back to the original name. Then create a blank migration inside your app and recreate the sql commands you did in the shell inside that migration file. That way django can keep track of the database schema.
You should change the name of the table in models.py not in MySQL shell.
From
class MyModel(models.Model):
...
To
class ThisModel(models.Model):
...
Or Create Proxy Model :
class ThisModel(MyModel):
class Meta:
proxy = True
verbose_name = "ThisModel"
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
I'm learning Django from Tango with Django but I keep getting this error when I type:
python manage.py makemigrations rango
python manage.py migrate
This is the output:
django.db.utils.IntegrityError: UNIQUE constraint failed: rango_category__new.slug
Models.py:
from django.db import models
from django.template.defaultfilters import slugify
class Category(models.Model):
name = models.CharField(max_length=128, unique=True)
views = models.IntegerField(default=0)
likes = models.IntegerField(default=0)
slug = models.SlugField(unique=True)
def save(self, *args, **kwargs):
self.slug = slugify(self.name)
super(Category, self).save(*args, **kwargs)
def __unicode__(self):
return self.name
class Page(models.Model):
category = models.ForeignKey(Category)
title = models.CharField(max_length=128)
url = models.URLField()
views = models.IntegerField(default=0)
def __unicode__(self):
return self.title
The reason for this constrain could be that you didn't have any field called slug in Category class when you have initially migrated it (First Migration), and after adding this field in the model, when you ran makemigrations, you have set default value to something static value(i.e None or '' etc), and which broke the unique constrain for the Category's table's slug column in which slug should be unique but it isn't because all the entry will get that default value.
To solve this, you can either drop the database and migration files and re-run makemigrations and migrate or set a unique default value like this:
slug = models.SlugField(unique=True, default=uuid.uuid1)
Edit:
According to Migrations that add unique fields, modify your migration file to overcome unique constrain. For example, modify your migration file (which added the slug field to the model) like this:
import uuid
from app.models import Category # where app == tango_app_name
class Migration(migrations.Migration):
dependencies = [
('yourproject', '0003_remove_category_slug'),
]
def gen_uuid(apps, schema_editor):
for row in Category.objects.all():
row.slug = uuid.uuid4()
row.save()
operations = [
migrations.AddField(
model_name='category',
name='slug',
field=models.SlugField(default=uuid.uuid4),
preserve_default=True,
),
migrations.RunPython(gen_uuid),
migrations.AlterField(
model_name='category',
name='slug',
field=models.SlugField(default=uuid.uuid4, unique=True),
),
]
I got a field with attribute unique, which was not unique [eg 2-time same value]
python3 manage.py migrate --fake
then
python3 manage.py makemigrations
python3 manage.py migrate
this did the trick
This means a slug should be unique. You may have some data in your model. You need to delete all the data in that model and you need to migrate again.
In this situation, you have two ways to fix the error;
You need to delete it from the Django admin site. More often than not, it may give an error when you are trying to open the model.
Open command prompt
move to project -> py manage.py shell -> from yourappname.models import modelname -> modelname.objects.delete()
Here if you define a product manager for your model. Then you have to define a delete function. Later you should makemigrate, migrate and continue with the second way
I just met this simiilar error: Django UNIQUE constraint failed. I tried examine the code for very long time, but didn't solve it. I finally used SQLiteStudio to examine the data, and found the data is problematic: I unintentionally added two SAME instances which violates the UNIQUE constraint. To be frank I haven't thought the error could be this naive and simple, and because so it took me a lot of time to find out!
I had the same problem and tried all the suggested answers. What finally worked for me was, after I defined the slug field as a URL in models, and ran the makemigrations. I edited the file in makemigrations adding a random number at the end of a basic URL, like this
Generated by Django 3.2.3 on 2022-02-02 20:58
from django.db import migrations, models
from random import randint
class Migration(migrations.Migration):
dependencies = [
('blog', '0002_remove_post_slug1'),
]
operations = [
migrations.AddField(
model_name='post',
name='slug',
field=models.URLField(blank=True, default='http:/salpimientapa.com/' + str(randint(100000,999999))),
),
]
After I ran
python manage.py migrate
I edit the slug as a SlugModel and ran the makemigrations and migrate again
What worked for me was going to the admin and changing the value of duplicate slug, before running the migrations again.
Just delete the last migration in the migration folder
Then run
python manage.py makemigrations
python manage.py migrate
I faced the same issue and solved by populating my slugfied thro' the admin with unique values and without leaving any of them blank.
Basically: You add the field without unique=true in one operation, make a data migration that generates the correct shortuuids for you, and then change the field too unique again.
i have this error too ,
i did delete my database in djangoProject ( for example db.sqlite3 )
and then run
python manage.py makemigrations
python manage.py migrate
It's an Integrity Error probably because the migration will temper with the already exiting data in the database.
I had this error and here's what I did:
Enter in the project folder directory
Open the python interpreter
py manage.py shell
Import your Models
from yourappname.models import model
Delete existing data records in the model
model.objects.all().delete()
Exit the Python Interpreter
exit()
.
Another thing you could do is to set unique="false" on the affecting field. I think this should work; not so sure.
Am trying to create simple blog using django.
At first,i created database with the command
python manage.py syncdb
when i try to save blog post,i get the following error
DatabaseError: table blog_app_post has no column named body
models.py code :
from django.db import models
from taggit.managers import TaggableManager
class Post(models.Model):
title = models.CharField(max_length=255)
body = models.TextField()
created = models.DateTimeField()
tags = TaggableManager()
def __unicode__(self):
return self.title
but the column named body is actually created in the Db.
BEGIN;
CREATE TABLE "blog_app_post" (
"id" integer NOT NULL PRIMARY KEY,
"title" varchar(255) NOT NULL,
"body" text NOT NULL,
"created" datetime NOT NULL
)
what does this error mean and anyone would propose a solution for this?
This is probably because you changed the structure of your posts data structure. What you need to do now is delete the schema for your previous table and paste in the new one.
You can avoid problems like this by using migration managers like south.
So, in order to solve this, run manage.py sql <app_name>, then you simply copy the latest SQL table on the list, the first one that is printed. Then you simply maange.py dbshell and then just paste and run the SQL.
How do you say that it's created, you checking it using python manage.py sqlall?
Did you add field body after running syncdb initially. In that case you will have to use a migration.
so i'm trying to do a data migration where i take the "listings" from a realestate app into a new "listings" app that i've created.
i did startmigration like this:
python manage.py startmigration listings migrate_listings --freeze realestate
created a blank migration, which i populated with this:
def forwards(self, orm):
"Write your forwards migration here"
for listing in orm['realestate.RealEstateListing'].objects.all():
sub_type = orm.SubType.objects.get(slug_url=slugify(listing.listing_type.name))
lt = orm.Listing(listing_type=sub_type.parent,
sub_type=sub_type,
expiration_date=listing.expiration_date,
title=listing.title,
slug_url = listing.slug_url,
description = listing.description,
contact_person=listing.contact_person,
secondary_contact=listing.secondary_contact,
address=listing.address,
location=listing.location,
price=listing.price,
pricing_option=listing.pricing_option,
display_picture=listing.display_picture,
image_gallery=listing.image_gallery,
date_added=listing.date_added,
status=listing.status,
featured_on_homepage=listing.featured_on_homepage,
)
lt.save()
lt.features.clear()
for ft in listing.property_features.all:
lt.features.add(ft)
for cft in listing.community_features.all:
lt.features.add(cft)
lt.restrictions.clear()
for na in listing.not_allowed.all:
lt.restrictions.add(na)
however when i run the migration is still get this error:
whiney_method
ValueError("you cannot instantiate a stub model")
from what i understand you can't access a "stub" model using the fakeorm but freezing additional apps is not allowed. how do i go about using the "stub" models without freezing them?
ok so i'm answering my own question, since apparantly i'm the only django south user here. i had to figure it out by myself.
what i wasn't doing, was freezing all the apps that were required in the above migration. since i didn't freeze it created the stub models.
the proper syntax for freezing multiple apps is:
python manage.py startmigration listings migrate_listings --freeze realestate --freeze logistics --freeze media --freeze upload
and everything works after that!