Django Newbie: python manage.py syncdb error? - django

Question

Working on the Django tutorial at the moment and got this error when I tried:
python manage.py syncdb
Using sqlite3
Traceback (most recent call last):
File "manage.py", line 10, in <module>
execute_from_command_line(sys.argv)
File "/Library/Python/2.6/site-packages/django/core/management/__init__.py", line 443, in execute_from_command_line
utility.execute()
File "/Library/Python/2.6/site-packages/django/core/management/__init__.py", line 382, in execute
self.fetch_command(subcommand).run_from_argv(self.argv)
File "/Library/Python/2.6/site-packages/django/core/management/base.py", line 196, in run_from_argv
self.execute(*args, **options.__dict__)
File "/Library/Python/2.6/site-packages/django/core/management/base.py", line 232, in execute
output = self.handle(*args, **options)
File "/Library/Python/2.6/site-packages/django/core/management/base.py", line 371, in handle
return self.handle_noargs(**options)
File "/Library/Python/2.6/site-packages/django/core/management/commands/syncdb.py", line 57, in handle_noargs
cursor = connection.cursor()
File "/Library/Python/2.6/site-packages/django/db/backends/__init__.py", line 306, in cursor
cursor = self.make_debug_cursor(self._cursor())
File "/Library/Python/2.6/site-packages/django/db/backends/sqlite3/base.py", line 281, in _cursor
self._sqlite_create_connection()
File "/Library/Python/2.6/site-packages/django/db/backends/sqlite3/base.py", line 271, in _sqlite_create_connection
self.connection = Database.connect(**kwargs)
sqlite3.OperationalError: unable to open database file
My settings.py file reads:
'ENGINE': 'django.db.backends.sqlite3', # Add 'postgresql_psycopg2', 'mysql', 'sqlite3' or 'oracle'.
'NAME': '/Users/username/django_tutorial/mysite',
Anyone know the problem here? it seems like I am putting in the complete file directory, but the databases still couldn't be created? ):

Answer 1

You should use the path to the file, including the file name:
'NAME': '/Users/username/django_tutorial/mysite/data.sqlite',
You can also just give it the file name:
'NAME': 'data.sqlite',
In that case it will be created in the same directory than the one from which you run manage.py.
The best practice is to pass an absolute path, but to ensure this absolute path is created relatively to the settings file. E.G:
import os
SETTINGS_DIR = os.path.dirname(os.path.abspath(__file__))
Then latter:
'NAME': os.path.join(SETTINGS_DIR, 'data.sqlite'),
Remember that the settings file is a Python file, and that you can use Python code in it.

Answer 2

There is a list of Django newbie mistakes on project's wiki. See this entry: https://code.djangoproject.com/wiki/NewbieMistakes#DjangosaysUnabletoOpenDatabaseFilewhenusingSQLite3
This specific error is common mistake among Django's newbies. See detailed description below.
PROBLEM
You're using SQLite3, your DATABASE_NAME is set to the
database file's full path, the database file is writeable by Apache,
but you still get the above error.
SOLUTION
Make sure Apache can also write to the parent directory of
the database. SQLite needs to be able to write to this directory.
Make sure each folder of your database file's full path does not start
with number, eg. /www/4myweb/db (observed on Windows 2000).
If DATABASE_NAME is set to something like
'/Users/yourname/Sites/mydjangoproject/db/db', make sure you've
created the 'db' directory first.
Make sure your /tmp directory is world-writable (an unlikely cause as
other thing on your system will also not work). ls /tmp -ald should
produce drwxrwxrwt ....
Make sure the path to the database specified in settings.py is a full
path.
If you working on windows make also sure that you have the path to the
db directory written with double backlashes
'C:\\django\\sqlite\\django.db'

Answer 3

i dont think you need the full path for the name, just put the file name ie "my_db.db". that is assuming you have sqlite3 installed.