How to fix error while working with Django in Pycharm?

Question

Django

How to fix a database error?
When I run the syncdb command , this is what happens:

"C:\Program Files (x86)\JetBrains\PyCharm 3.4.1\bin\runnerw.exe" C:\Python26\python.exe "C:\Program Files (x86)\JetBrains\PyCharm 3.4.1\helpers\pycharm\django_manage.py" syncdb C:/Users/ТВ/PycharmProjects/djblog
Traceback (most recent call last):
  File "C:\Program Files (x86)\JetBrains\PyCharm 3.4.1\helpers\pycharm\django_manage.py", line 23, in <module>
    run_module(manage_file, None, '__main__', True)
  File "C:\Python26\lib\runpy.py", line 136, in run_module
    fname, loader, pkg_name)
  File "C:\Python26\lib\runpy.py", line 54, in _run_module_code
    mod_loader, pkg_name)
  File "C:\Python26\lib\runpy.py", line 34, in _run_code
    exec code in run_globals
  File "C:\Users\��\PycharmProjects\djblog\manage.py", line 10, in <module>
    execute_from_command_line(sys.argv)
  File "C:\Python26\lib\site-packages\django\core\management\__init__.py", line 399, in execute_from_command_line
    utility.execute()
  File "C:\Python26\lib\site-packages\django\core\management\__init__.py", line 392, in execute
    self.fetch_command(subcommand).run_from_argv(self.argv)
  File "C:\Python26\lib\site-packages\django\core\management\base.py", line 242, in run_from_argv
    self.execute(*args, **options.__dict__)
  File "C:\Python26\lib\site-packages\django\core\management\base.py", line 285, in execute
    output = self.handle(*args, **options)
  File "C:\Python26\lib\site-packages\django\core\management\base.py", line 415, in handle
    return self.handle_noargs(**options)
  File "C:\Python26\lib\site-packages\django\core\management\commands\syncdb.py", line 57, in handle_noargs
    cursor = connection.cursor()
  File "C:\Python26\lib\site-packages\django\db\backends\__init__.py", line 160, in cursor
    cursor = self.make_debug_cursor(self._cursor())
  File "C:\Python26\lib\site-packages\django\db\backends\__init__.py", line 132, in _cursor
    self.ensure_connection()
  File "C:\Python26\lib\site-packages\django\db\backends\__init__.py", line 127, in ensure_connection
    self.connect()
  File "C:\Python26\lib\site-packages\django\db\utils.py", line 99, in __exit__
    six.reraise(dj_exc_type, dj_exc_value, traceback)
  File "C:\Python26\lib\site-packages\django\db\backends\__init__.py", line 127, in ensure_connection
    self.connect()
  File "C:\Python26\lib\site-packages\django\db\backends\__init__.py", line 115, in connect
    self.connection = self.get_new_connection(conn_params)
  File "C:\Python26\lib\site-packages\django\db\backends\sqlite3\base.py", line 346, in get_new_connection
    conn = Database.connect(**conn_params)
django.db.utils.OperationalError: unable to open database file

Although the database file is in its place in the project folder. What's wrong?

Answer 1

I fixed everything, just in the path to the project folder there was a folder with Cyrillic, stuffed it into the root of the disk and everything is fine)

Answer 2

You don't need to create it. Delete and execute...
Which synсdb? makemigrations && migrate