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Django TypeError: id() takes exactly one argument (0 given)

So I have been trying to implement a way to upload multiple images to a post. The way I did it is to have tables. One for the actual post, and one of the multiple images uploaded. I was planning to link them with a foreign key but it is not working. My terminal started throwing the error "TypeError: id() takes exactly one argument (0 given)" . It throws me this error whenever I migrate it.
I am not sure how to fix this.
MY code:
models.py
from django.db import models
from django.utils import timezone
from django.forms import ModelForm
from django.utils.text import slugify
from django.utils.crypto import get_random_string
from django.conf import settings
from PIL import Image
import os
DEFAULT_IMAGE_ID = 1
# Create your models here.
class Projects(models.Model):
title = models.CharField(max_length=30)
description = models.TextField(max_length=150)
publish_date = models.DateTimeField(auto_now=False, auto_now_add=True)
update_date = models.DateTimeField(auto_now=True, auto_now_add=False)
slug = models.SlugField(unique=True)
files = models.FileField(upload_to='files/', blank=True)
images = models.ImageField(upload_to='images/', height_field = 'img_height', width_field = 'img_width',blank=True)
img_height = models.PositiveIntegerField(default=600)
img_width = models.PositiveIntegerField(default=300)
#feature_images = models.ForeignKey(P_Images, on_delete=models.CASCADE, default=DEFAULT_IMAGE_ID)
feature_images = models.AutoField(primary_key=True)
def __str__(self):
return self.title
def save(self, *args, **kwargs):
# Generates a random string
unique_string = get_random_string(length=32)
# Combines title and unique string to slugify
slugtext = self.title + "-" + "unique_id=-" + unique_string
self.slug = slugify(slugtext)
return super(Projects, self).save(*args, **kwargs)
class P_Images(models.Model):
p_file = models.ImageField(upload_to='images/', blank=None)
p_uploaded_at = models.DateTimeField(auto_now_add=True, auto_now=False)
#fk_post = models
fk_post = models.ForeignKey(Projects, on_delete=models.CASCADE)
The errorlog
Operations to perform:
Apply all migrations: admin, auth, contenttypes, projects, sessions
Running migrations:
Applying projects.0005_auto_20180823_0553...Traceback (most recent call last):
File "manage.py", line 15, in <module>
execute_from_command_line(sys.argv)
File "/home/erichardson/env01/lib/python3.6/site-packages/django/core/management/__init__.py", line 371, in execute_from_command_line
utility.execute()
File "/home/erichardson/env01/lib/python3.6/site-packages/django/core/management/__init__.py", line 365, in execute
self.fetch_command(subcommand).run_from_argv(self.argv)
File "/home/erichardson/env01/lib/python3.6/site-packages/django/core/management/base.py", line 288, in run_from_argv
self.execute(*args, **cmd_options)
File "/home/erichardson/env01/lib/python3.6/site-packages/django/core/management/base.py", line 335, in execute
output = self.handle(*args, **options)
File "/home/erichardson/env01/lib/python3.6/site-packages/django/core/management/commands/migrate.py", line 200, in handle
fake_initial=fake_initial,
File "/home/erichardson/env01/lib/python3.6/site-packages/django/db/migrations/executor.py", line 117, in migrate
state = self._migrate_all_forwards(state, plan, full_plan, fake=fake, fake_initial=fake_initial)
File "/home/erichardson/env01/lib/python3.6/site-packages/django/db/migrations/executor.py", line 147, in _migrate_all_forwards
state = self.apply_migration(state, migration, fake=fake, fake_initial=fake_initial)
File "/home/erichardson/env01/lib/python3.6/site-packages/django/db/migrations/executor.py", line 244, in apply_migration
state = migration.apply(state, schema_editor)
File "/home/erichardson/env01/lib/python3.6/site-packages/django/db/migrations/migration.py", line 122, in apply
operation.database_forwards(self.app_label, schema_editor, old_state, project_state)
File "/home/erichardson/env01/lib/python3.6/site-packages/django/db/migrations/operations/fields.py", line 216, in database_forwards
schema_editor.alter_field(from_model, from_field, to_field)
File "/home/erichardson/env01/lib/python3.6/site-packages/django/db/backends/base/schema.py", line 525, in alter_field
old_db_params, new_db_params, strict)
File "/home/erichardson/env01/lib/python3.6/site-packages/django/db/backends/base/schema.py", line 630, in _alter_field
new_default = self.effective_default(new_field)
File "/home/erichardson/env01/lib/python3.6/site-packages/django/db/backends/base/schema.py", line 218, in effective_default
default = field.get_default()
File "/home/erichardson/env01/lib/python3.6/site-packages/django/db/models/fields/__init__.py", line 775, in get_default
return self._get_default()
TypeError: id() takes exactly one argument (0 given)
I use a MySQL database for this. This error started popping up after I updated my tables to be able to link with each other. I plan the fk_post of the P_Images to contain the feature_image value of Projects for the foreign key.
005_migration.py
import builtins
from django.db import migrations, models
import django.db.models.deletion
class Migration(migrations.Migration):
dependencies = [
('projects', '0004_auto_20180823_0547'),
]
operations = [
migrations.AlterField(
model_name='p_images',
name='fk_post',
field=models.ForeignKey(on_delete=django.db.models.deletion.CASCADE, to='projects.Projects'),
),
migrations.AlterField(
model_name='projects',
name='feature_images',
field=models.IntegerField(default=builtins.id),
),
]
please let me know if the migration.py tells you something or nothing.

You have something very bizarre in your migration:
models.IntegerField(default=builtins.id)
This is referring to the builtin id function, which requires an argument because it returns the internal ID of an object in Python. It has nothing to do with database IDs, and doesn't belong here at all. I can only guess that you were asked for a default when creating the migration and you just typed in id.
You should delete that default from the migration, but that may make it unable to execute. You could also try a default of 0, which makes sense there; but your actual models code shows that field as the primary key, so you presumably have another subsequent migration that changes the field again; and 0 wouldn't work as a pk.
If you're still in development and don't have any data you need to keep, I would suggest deleting your database and migrations completely and starting again with makemigrations.

I think this error occurred because you haven't set a default value to your foreign key. For your previous P_Images instances (before your current migration), they don't have a Projects id related to them.
The error text shown also confirms it:
return self._get_default()
TypeError: id() takes exactly one argument (0 given)
So try adding a default value :
DEFAULT_PROJECT = 1 # or the id of any project that does exist
fk_post = models.ForeignKey(Projects, on_delete=models.CASCADE, default = DEFAULT_PROJECT)

ValueError on foreignkey default value

I deleted my db.sqlite3 and the file on my migration folder except __init__.py but including the __pycache__ folder.
My app's models are multiple files in a folder with an __init__.py file with some import and an __all__ variable (I don't know if this matter).
From the prompt I give the first migrate command and it look to work (migrating sessions, admin, auth, registration, contenttypes. There isn't my app 'core').
So i give the first makemigrations command and again the migrate command.
This time it give an error:
ValueError: invalid literal for int() with base 10: 'basic'
In a model there's an ForeignKey fields with default='basic'. basic will be an object of the related class but actually it don't exist (before I create the database, then I will populate it). I think it's normal.
Anyway I change in default=1 and it works (but 1 will not be an object, so it's wrong!)
My model:
Class Payment(models.Model):
name = models.CharField(max_length=32)
Class ProfileUser(models.Model):
payment = models.ForeignKey(Payment, blank=True, null=True,
default='basic', on_delete=models.PROTECT)
#etc
There is a way to use 'basic'? I prefer to not use id=1 because I'm not sure to know the id that basic will have (remember that now it don't exist yet).
Btw in other models I have alike situations and they appear to works...
The Whole error:
(myvenv) c:\Python34\Scripts\possedimenti\sitopossedimenti>manage.py migrate
Operations to perform:
Apply all migrations: contenttypes, core, registration, admin, auth, sessions
Running migrations:
Rendering model states... DONE
Applying core.0001_initial...Traceback (most recent call last):
File "C:\Python34\Scripts\possedimenti\sitopossedimenti\manage.py", line 10, i
n <module>
execute_from_command_line(sys.argv)
File "c:\Python34\Scripts\possedimenti\myvenv\lib\site-packages\django\core\ma
nagement\__init__.py", line 353, in execute_from_command_line
utility.execute()
File "c:\Python34\Scripts\possedimenti\myvenv\lib\site-packages\django\core\ma
nagement\__init__.py", line 345, in execute
self.fetch_command(subcommand).run_from_argv(self.argv)
File "c:\Python34\Scripts\possedimenti\myvenv\lib\site-packages\django\core\ma
nagement\base.py", line 348, in run_from_argv
self.execute(*args, **cmd_options)
File "c:\Python34\Scripts\possedimenti\myvenv\lib\site-packages\django\core\ma
nagement\base.py", line 399, in execute
output = self.handle(*args, **options)
File "c:\Python34\Scripts\possedimenti\myvenv\lib\site-packages\django\core\ma
nagement\commands\migrate.py", line 200, in handle
executor.migrate(targets, plan, fake=fake, fake_initial=fake_initial)
File "c:\Python34\Scripts\possedimenti\myvenv\lib\site-packages\django\db\migr
ations\executor.py", line 92, in migrate
self._migrate_all_forwards(plan, full_plan, fake=fake, fake_initial=fake_ini
tial)
File "c:\Python34\Scripts\possedimenti\myvenv\lib\site-packages\django\db\migr
ations\executor.py", line 121, in _migrate_all_forwards
state = self.apply_migration(state, migration, fake=fake, fake_initial=fake_
initial)
File "c:\Python34\Scripts\possedimenti\myvenv\lib\site-packages\django\db\migr
ations\executor.py", line 198, in apply_migration
state = migration.apply(state, schema_editor)
File "c:\Python34\Scripts\possedimenti\myvenv\lib\site-packages\django\db\migr
ations\migration.py", line 123, in apply
operation.database_forwards(self.app_label, schema_editor, old_state, projec
t_state)
File "c:\Python34\Scripts\possedimenti\myvenv\lib\site-packages\django\db\migr
ations\operations\fields.py", line 62, in database_forwards
field,
File "c:\Python34\Scripts\possedimenti\myvenv\lib\site-packages\django\db\back
ends\sqlite3\schema.py", line 221, in add_field
self._remake_table(model, create_fields=[field])
File "c:\Python34\Scripts\possedimenti\myvenv\lib\site-packages\django\db\back
ends\sqlite3\schema.py", line 103, in _remake_table
self.effective_default(field)
File "c:\Python34\Scripts\possedimenti\myvenv\lib\site-packages\django\db\back
ends\base\schema.py", line 210, in effective_default
default = field.get_db_prep_save(default, self.connection)
File "c:\Python34\Scripts\possedimenti\myvenv\lib\site-packages\django\db\mode
ls\fields\related.py", line 912, in get_db_prep_save
return self.target_field.get_db_prep_save(value, connection=connection)
File "c:\Python34\Scripts\possedimenti\myvenv\lib\site-packages\django\db\mode
ls\fields\__init__.py", line 728, in get_db_prep_save
prepared=False)
File "c:\Python34\Scripts\possedimenti\myvenv\lib\site-packages\django\db\mode
ls\fields\__init__.py", line 968, in get_db_prep_value
value = self.get_prep_value(value)
File "c:\Python34\Scripts\possedimenti\myvenv\lib\site-packages\django\db\mode
ls\fields\__init__.py", line 976, in get_prep_value
return int(value)
ValueError: invalid literal for int() with base 10: 'basic'
Thank You
PS: I'm using python 3.4 Django 1.9 Windows7

In your Payment model you have not explicitly designated a primary key. So django will create one for you. It will be an integer auto increment field.
Class Payment(models.Model):
name = models.CharField(max_length=32)
Now when you define Payment as a foreign key for ProfileUser, django thinks that it should be the primary key field in Payment that ought to be used as the reference.
Class ProfileUser(models.Model):
payment = models.ForeignKey(Payment, blank=True, null=True,
default='basic', on_delete=models.PROTECT)
#etc
In order to enforce this reference, the payment field needs to be an integer. And you cannot insert 'basic' into an integer field.
One solution is to use the
ForeignKey.to_field option to explicitly refer to a different field in the Payment class.
ForeignKey.to_field¶ The field on the related object that the relation
is to. By default, Django uses the primary key of the related object.
Thus your class would become
Class ProfileUser(models.Model):
payment = models.ForeignKey(Payment, blank=True, null=True,
default='basic', on_delete=models.PROTECT,
to_field = 'name' )
#etc
Now the payment field here becomes a CharField. Note that it would be more efficient to use an integer field as a foreign key in most cases.
Another possibility is for you to define the name field in Payment as it's primary key.

Create a function to return the id (put it before the ProfileUser defenition).
def get_basic_id():
payment = Payment.options.get(name='basic')
return payment.id
Class ProfileUser(models.Model):
payment = models.ForeignKey(Payment, blank=True, null=True,
default=get_basic_id(), on_delete=models.PROTECT)
Though if you don't know what id 'basic' will have, are you certain that it will be in the database? One way around this would be to populate the payment table using fixtures.
https://docs.djangoproject.com/en/1.9/howto/initial-data/

Django Models Relations to User Model

I'm new to the forum and I've got a problem.
in Django im trying to create a UserProfile class by referencing it via a OnetoOneField to an User Object. When i try to migrate, I get:
"You are trying to add a non-nullable field 'id' to author 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"
I understand, that there are some fields in the User model (for example "id") that are not allowed to be null.
When I try to set a default it will makemigrations, but migrate fails with a syntax error.
My Question is how do I allow Null, or set a default that works, without writing a custom User class, and is that even possible ?
I'd really appreciate any help with this.
Here is my Code:
from django.db import models
from django.contrib.auth.models import User as User_django
class User(User_django):
pass
class Author(models.Model):
user = models.OneToOneField(User, null=True)
class BaseElement(models.Model):
author=models.ForeignKey(Author, null=True)
upvotes = models.PositiveIntegerField(default=0)
downvotes = models.PositiveIntegerField(default=0)
created = models.DateTimeField(auto_now_add=True, null=True)
def vote_up(self):
self.upvotes.value = self.upvotes.value + 1
def vote_down(self):
self.downvotes.value = self.downvotes.value + 1
class Meta:
abstract = True
class Post(BaseElement):
content = models.TextField(max_length=240)
tags = models.CharField(max_length=40, blank=True)
said_by = models.CharField(max_length=40)
said_at = models.CharField(max_length=40, blank=True)
def get_comments(self):
return self.comment_set.all()
def get_comments_anzahl(self):
return self.comment_set.all().count()
class Comment(BaseElement):
content = models.TextField(max_length=140)
commented_post = models.ForeignKey(Post, related_query_name="comment", null=True)
EDIT:
This happens if I try to set a default and try to migrate:
You are trying to add a non-nullable field 'id' to author 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()
>>> 1
Migrations for 'post':
0003_auto_20160505_1237.py:
- Change Meta options on author
- Change managers on author
- Remove field user_ptr from author
- Add field id to author
- Add field user to author
Operations to perform:
Apply all migrations: contenttypes, admin, sessions, post, auth
Running migrations:
Rendering model states... DONE
Applying post.0003_auto_20160505_1237...Traceback (most recent call last):
File "/home/jaidmin/.conda/envs/cyf/lib/python3.5/site-packages/django/db/backends/utils.py", line 64, in execute
return self.cursor.execute(sql, params)
File "/home/jaidmin/.conda/envs/cyf/lib/python3.5/site-packages/django/db/backends/sqlite3/base.py", line 323, in execute
return Database.Cursor.execute(self, query, params)
sqlite3.OperationalError: near ")": syntax error
The above exception was the direct cause of the following exception:
Traceback (most recent call last):
File "manage.py", line 10, in <module>
execute_from_command_line(sys.argv)
File "/home/jaidmin/.conda/envs/cyf/lib/python3.5/site-packages/django /core/management/__init__.py", line 353, in execute_from_command_line
utility.execute()
File "/home/jaidmin/.conda/envs/cyf/lib/python3.5/site-packages/django/core/management/__init__.py", line 345, in execute
self.fetch_command(subcommand).run_from_argv(self.argv)
File "/home/jaidmin/.conda/envs/cyf/lib/python3.5/site-packages/django/core/management/base.py", line 348, in run_from_argv
self.execute(*args, **cmd_options)
File "/home/jaidmin/.conda/envs/cyf/lib/python3.5/site-packages/django/core/management/base.py", line 399, in execute
output = self.handle(*args, **options)
File "/home/jaidmin/.conda/envs/cyf/lib/python3.5/site-packages/django/core/management/commands/migrate.py", line 200, in handle
executor.migrate(targets, plan, fake=fake, fake_initial=fake_initial)
File "/home/jaidmin/.conda/envs/cyf/lib/python3.5/site-packages/django/db/migrations/executor.py", line 92, in migrate
self._migrate_all_forwards(plan, full_plan, fake=fake, fake_initial=fake_initial)
File "/home/jaidmin/.conda/envs/cyf/lib/python3.5/site-packages/django/db/migrations/executor.py", line 121, in _migrate_all_forwards
state = self.apply_migration(state, migration, fake=fake, fake_initial=fake_initial)
File "/home/jaidmin/.conda/envs/cyf/lib/python3.5/site-packages/django/db/migrations/executor.py", line 198, in apply_migration
state = migration.apply(state, schema_editor)
File "/home/jaidmin/.conda/envs/cyf/lib/python3.5/site-packages/django/db/migrations/migration.py", line 123, in apply
operation.database_forwards(self.app_label, schema_editor, old_state, project_state)
File "/home/jaidmin/.conda/envs/cyf/lib/python3.5/site-packages/django/db/migrations/operations/fields.py", line 121, in database_forwards
schema_editor.remove_field(from_model, from_model._meta.get_field(self.name))
File "/home/jaidmin/.conda/envs/cyf/lib/python3.5/site-packages/django/db/backends/sqlite3/schema.py", line 247, in remove_field
self._remake_table(model, delete_fields=[field])
File "/home/jaidmin/.conda/envs/cyf/lib/python3.5/site-packages/django/db/backends/sqlite3/schema.py", line 197, in _remake_table
self.quote_name(model._meta.db_table),
File "/home/jaidmin/.conda/envs/cyf/lib/python3.5/site-packages/django/db/backends/base/schema.py", line 110, in execute
cursor.execute(sql, params)
File "/home/jaidmin/.conda/envs/cyf/lib/python3.5/site-packages/django/db/backends/utils.py", line 79, in execute
return super(CursorDebugWrapper, self).execute(sql, params)
File "/home/jaidmin/.conda/envs/cyf/lib/python3.5/site-packages/django/db/backends/utils.py", line 64, in execute
return self.cursor.execute(sql, params)
File "/home/jaidmin/.conda/envs/cyf/lib/python3.5/site-packages/django/db/utils.py", line 95, in __exit__
six.reraise(dj_exc_type, dj_exc_value, traceback)
File "/home/jaidmin/.conda/envs/cyf/lib/python3.5/site-packages/django/utils/six.py", line 685, in reraise
raise value.with_traceback(tb)
File "/home/jaidmin/.conda/envs/cyf/lib/python3.5/site-packages/django/db/backends/utils.py", line 64, in execute
return self.cursor.execute(sql, params)
File "/home/jaidmin/.conda/envs/cyf/lib/python3.5/site-packages/django/db/backends/sqlite3/base.py", line 323, in execute
return Database.Cursor.execute(self, query, params)
django.db.utils.OperationalError: near ")": syntax error
EDIT:
It seems like I solved it by creating a new project and step by step copying the models and migrating. Still wondering what was going on there, because I flushed the database and deleted the migration files multiple times and it was still throwing errors.
Thanks for the help !

Take a look at this:
https://docs.djangoproject.com/es/1.9/topics/db/examples/one_to_one/
Your code is a ONE-to-ONE with user. When you set it, it implies CASCADE delete, and therefore can never be null.
if you change to One-to-Many, and just ensure that you only insert one object for the foreign key, it'll work. Otherwise, change your cascade rule.
I will say, however, that's not a good idea, because if you make it null, what separates it from other records. Consider a second PK field.
Best of luck.

Just import django's built-in User model like this:
from django.contrib.auth.models import User
and delete this line:
class User(User_django):
pass
from your code as it is completely unnecessary.
UPDATE: Also it looks like your problem has nothing to do with the User model. I think somehow you added an id field to your Author model and deleted it (maybe after an unsuccessful migration). If this is the case, then you could edit your 0003_auto_20160505_1237.py. To do that open it and remove the lines which look something like this:
migrations.AddField(
model_name='author',
name='id',
field=models.someFieldType(options)),
),
and try again.

Django Db Utils IntegrityError: (1062, "Duplicate entry '' for key 'slug'") [duplicate]

I'm trying to follow the tangowithdjango book and must add a slug to update the category table. However I'm getting an error after trying to migrate the databases.
http://www.tangowithdjango.com/book17/chapters/models_templates.html#creating-a-details-page
I didn't provide a default value for the slug, so Django asked me to provide one and as the book instructed I type in ''.
It's worth noticing that instead of using sqlite as in the original book I'm using mysql.
models.py
from django.db import models
from django.template.defaultfilters import slugify
# Create your models here.
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)
class Meta:
verbose_name_plural = "Categories"
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 command prompt
sudo python manage.py migrate
Operations to perform:
Apply all migrations: admin, rango, contenttypes, auth, sessions
Running migrations:
Applying rango.0003_category_slug...Traceback (most recent call last):
File "manage.py", line 10, in <module>
execute_from_command_line(sys.argv)
File "/usr/local/lib/python2.7/dist-packages/django/core/management/__init__.py", line 385, in execute_from_command_line
utility.execute()
File "/usr/local/lib/python2.7/dist-packages/django/core/management/__init__.py", line 377, in execute
self.fetch_command(subcommand).run_from_argv(self.argv)
File "/usr/local/lib/python2.7/dist-packages/django/core/management/base.py", line 288, in run_from_argv
self.execute(*args, **options.__dict__)
File "/usr/local/lib/python2.7/dist-packages/django/core/management/base.py", line 338, in execute
output = self.handle(*args, **options)
File "/usr/local/lib/python2.7/dist-packages/django/core/management/commands/migrate.py", line 160, in handle
executor.migrate(targets, plan, fake=options.get("fake", False))
File "/usr/local/lib/python2.7/dist-packages/django/db/migrations/executor.py", line 63, in migrate
self.apply_migration(migration, fake=fake)
File "/usr/local/lib/python2.7/dist-packages/django/db/migrations/executor.py", line 97, in apply_migration
migration.apply(project_state, schema_editor)
File "/usr/local/lib/python2.7/dist-packages/django/db/migrations/migration.py", line 107, in apply
operation.database_forwards(self.app_label, schema_editor, project_state, new_state)
File "/usr/local/lib/python2.7/dist-packages/django/db/migrations/operations/fields.py", line 37, in database_forwards
field,
File "/usr/local/lib/python2.7/dist-packages/django/db/backends/mysql/schema.py", line 42, in add_field
super(DatabaseSchemaEditor, self).add_field(model, field)
File "/usr/local/lib/python2.7/dist-packages/django/db/backends/schema.py", line 411, in add_field
self.execute(sql, params)
File "/usr/local/lib/python2.7/dist-packages/django/db/backends/schema.py", line 98, in execute
cursor.execute(sql, params)
File "/usr/local/lib/python2.7/dist-packages/django/db/backends/utils.py", line 81, in execute
return super(CursorDebugWrapper, self).execute(sql, params)
File "/usr/local/lib/python2.7/dist-packages/django/db/backends/utils.py", line 65, in execute
return self.cursor.execute(sql, params)
File "/usr/local/lib/python2.7/dist-packages/django/db/utils.py", line 94, in __exit__
six.reraise(dj_exc_type, dj_exc_value, traceback)
File "/usr/local/lib/python2.7/dist-packages/django/db/backends/utils.py", line 65, in execute
return self.cursor.execute(sql, params)
File "/usr/local/lib/python2.7/dist-packages/django/db/backends/mysql/base.py", line 128, in execute
return self.cursor.execute(query, args)
File "/usr/local/lib/python2.7/dist-packages/MySQLdb/cursors.py", line 205, in execute
self.errorhandler(self, exc, value)
File "/usr/local/lib/python2.7/dist-packages/MySQLdb/connections.py", line 36, in defaulterrorhandler
raise errorclass, errorvalue
django.db.utils.IntegrityError: (1062, "Duplicate entry '' for key 'slug'")

Let's analyse it step by step:
You're adding slug field with unique = True, that means: each record must have different value, there can't be two records with same value in slug
You're creating migration: django asks you for default value for fields that exists already in database, so you provided '' (empty string) as that value.
Now django is trying to migrate your database. In database we have at least 2 records
First record is migrated, slug column is populated with empty string. That's good because no other record is having empty string in slug field
Second record is migrated, slug column is populated with empty string. That fails, because first record already have empty string in slug field. Exception is raised and migration is aborted.
That's why your migration fails. All you should do is to edit migration, copy migrations.AlterField operation twice, in first operation remove unique=True. Between that operations you should put migrations.RunPython operation and provide 2 parameters into that: generate_slugs and migrations.RunPython.noop.
Now you must create inside your migration function BEFORE migration class, name that function generate_slugs. Function should take 2 arguments: apps and schema_editor. In your function put at first line:
Category = apps.get_model('your_app_name', 'Category')
and now use Category.objects.all() to loop all your records and provide unique slug for each of them.

If you have more than one category in your table, then you cannot have unique=True and default='', because then you will have more than one category with slug=''. If your tutorial says to do this, then it's bad advice, although it might work in SQLite.
The correct approach to add a unique field to a model is:
Delete your current migration that isn't working.
Add the slug field, with unique=False. Create a new migration and run it.
Set a unique slug for every category. It sounds like the rango populate script might do this. Alternatively, you could write a migration to set the slugs, or even set them manually in the Django admin.
Change the slug field to unique=True. Create a new migration and run it.
If that's too difficult, then you could delete all your categories from your database except one. Then your current migration will run without having problems with the unique constraint. You can add the categories again afterwards.

You must have rows in your table already with empty slugs, which is a violation of the mysql unique constraint you created. You can update them manually by running manage.py dbshell to get to the mysql client, then updating the offending rows, e.g.
update table rango_category set slug = name where slug = '';
(assuming the rows with blank slugs have names). Or you can delete the rows with
delete from rango_category where slug = '';
After that, you should be able to run your migrations.

"unknown column X.id" error in django using existing DB

I am trying to create a model for an existsing DB. Using the output of manage.py inspectdb, My models.py file looks like this:
from django.db import models
...some more stuff here...
class Scripts(models.Model):
run_site = models.ForeignKey(Sites, db_column='run_site')
script_name = models.CharField(max_length=120)
module_name = models.CharField(unique=True, max_length=120)
type = models.CharField(max_length=24)
cat_name = models.CharField(max_length=90)
owner = models.ForeignKey(QAPeople, db_column='owner')
only_server = models.CharField(max_length=120, blank=True)
guest = models.IntegerField()
registered = models.IntegerField()
super = models.IntegerField()
admin = models.IntegerField()
run_timing = models.CharField(max_length=27)
manual_owner = models.ForeignKey(QAPeople, db_column='manual_owner')
script_id = models.IntegerField(unique=True,)
version = models.IntegerField()
comment = models.ForeignKey('ScriptComments', null=True, blank=True)
class Meta:
db_table = u'scripts'
When I try to do Scripts.objects.all() I get
Traceback (most recent call last):
File "<console>", line 1, in <module>
File "C:\Python26\lib\site-packages\django\db\models\query.py", line 68, in __repr__
data = list(self[:REPR_OUTPUT_SIZE + 1])
File "C:\Python26\lib\site-packages\django\db\models\query.py", line 83, in __len__
self._result_cache.extend(list(self._iter))
File "C:\Python26\lib\site-packages\django\db\models\query.py", line 269, in iterator
for row in compiler.results_iter():
File "C:\Python26\lib\site-packages\django\db\models\sql\compiler.py", line 672, in results_iter
for rows in self.execute_sql(MULTI):
File "C:\Python26\lib\site-packages\django\db\models\sql\compiler.py", line 727, in execute_sql
cursor.execute(sql, params)
File "C:\Python26\lib\site-packages\django\db\backends\util.py", line 15, in execute
return self.cursor.execute(sql, params)
File "C:\Python26\lib\site-packages\django\db\backends\mysql\base.py", line 86, in execute
return self.cursor.execute(query, args)
File "C:\Python26\lib\site-packages\MySQLdb\cursors.py", line 173, in execute
self.errorhandler(self, exc, value)
File "C:\Python26\lib\site-packages\MySQLdb\connections.py", line 36, in defaulterrorhandler
raise errorclass, errorvalue
OperationalError: (1054, "Unknown column 'scripts.id' in 'field list'")
Why does django think there should be a scripts.id column? How do I fix it without dropping the tables etc?

There is always by default an implicit id field as auto incrementing primary key on every model. See primary_key in the Django docs how to change that field to some other name, but there needs to be one primary key (also in your table).
Also you may not want to call one of your fields super, since it is shadowing Python's built-in super in the class body. Might give you a hard time finding a bug some day.

I'm not sure if you'r familiar with django-south?
It's a nice little tool that helps you migrate your data from one structure to another and also helps identify issues before commiting info to the table, causing less errors and debugging afterwards.