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using Django Sqlite in windows

开发者 https://www.devze.com 2023-04-12 07:55 出处:网络
I want to use sqlite for 开发者_Go百科development in django, with these settings: \'default\': { \'ENGINE\': \'django.db.backends.sqlite3\', # Add \'postgresql_psycopg2\', \'postgresql\', \'mysql\',

I want to use sqlite for 开发者_Go百科development in django, with these settings:

'default': {
    'ENGINE': 'django.db.backends.sqlite3', # Add 'postgresql_psycopg2', 'postgresql', 'mysql', 'sqlite3' or 'oracle'.
    'NAME': 'D:\database.db',                      # Or path to database file if using sqlite3.
    'USER': '',                      # Not used with sqlite3.
    'PASSWORD': '',                  # Not used with sqlite3.
    'HOST': '',                      # Set to empty string for localhost. Not used with sqlite3.
    'PORT': '',                      # Set to empty string for default. Not used with sqlite3.
}

but when i try to do syncdb i get the following:

Traceback (most recent call last):
File "manage.py", line 11, in <module>
execute_manager(settings)
File "C:\Python27\lib\site-packages\django\core\management\__init__.py", line
438, in execute_manager
utility.execute()
File "C:\Python27\lib\site-packages\django\core\management\__init__.py", line
379, in execute
self.fetch_command(subcommand).run_from_argv(self.argv)
File "C:\Python27\lib\site-packages\django\core\management\base.py", line 191,
in run_from_argv
self.execute(*args, **options.__dict__)
File "C:\Python27\lib\site-packages\django\core\management\base.py", line 220,
in execute
output = self.handle(*args, **options)
File "C:\Python27\lib\site-packages\django\core\management\base.py", line 351,
in handle
return self.handle_noargs(**options)
File "C:\Python27\lib\site-packages\django\core\management\commands\syncdb.py"
, line 56, in handle_noargs
cursor = connection.cursor()
File "C:\Python27\lib\site-packages\django\db\backends\__init__.py", line 250,
in cursor
cursor = self.make_debug_cursor(self._cursor())
File "C:\Python27\lib\site-packages\django\db\backends\sqlite3\base.py", line
207, in _cursor
self.connection = Database.connect(**kwargs)
sqlite3.OperationalError: unable to open database file

Any idea what's going on? Notice that I tried to change the db name (location) many times but it didn't work ...


Use forward slashes or r'' strings to begin with.

'default': {
    # [... snip ...]
    'NAME': 'D:/database.db',
    # (or...) 'NAME': r'D:\database.db',
    # [... snip ...]
}

Additionally you need to ensure the entire directory your Sqlite file is in is writable (in this case D:/), as Sqlite needs to write its journal file next to the database file.


This works for me

DATABASES = {
    'default': {
        'ENGINE': 'django.db.backends.sqlite3', 
        'NAME': 'c:/projects/blog/first.db',  
        'USER': '',                      # Not used with sqlite3.
        'PASSWORD': '',                  # Not used with sqlite3.
        'HOST': '',                      # Not used with sqlite3.
        'PORT': '',                      # Not used with sqlite3.
    }
}


Try using only the name of the database file (no path). By default, Django will place it in the same directory as manage.py

'default': {
    'ENGINE': 'django.db.backends.sqlite3', # Add 'postgresql_psycopg2', 'postgresql', 'mysql', 'sqlite3' or 'oracle'.
    'NAME': 'database.db',                      # Or path to database file if using sqlite3.
    'USER': '',                      # Not used with sqlite3.
    'PASSWORD': '',                  # Not used with sqlite3.
    'HOST': '',                      # Set to empty string for localhost. Not used with sqlite3.
    'PORT': '',                      # Set to empty string for default. Not used with sqlite3.
}
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