我已经建立了一个新的项目,并用简单的模型填充它。(本质上我是在关注图坦。)
当我在命令行上运行python manage.py shell
时,它运行得很好:
>python manage.py shell
Python 2.6.4 (r264:75708, Oct 26 2009, 08:23:19) [MSC v.1500 32 bit (Intel)] on
win32
Type "help", "copyright", "credits" or "license" for more information.
(InteractiveConsole)
>>> from mysite.myapp.models import School
>>> School.objects.all()
[]
效果很好。然后,我尝试在Eclipse中做同样的事情(使用由相同文件组成的Django项目)。
右键单击带有Django环境的mysite项目>> Django >> Shell
这是来自PyDev控制台的输出:
>>> import sys; print('%s %s' % (sys.executable or sys.platform, sys.version))
C:\Python26\python.exe 2.6.4 (r264:75708, Oct 26 2009, 08:23:19) [MSC v.1500 32 bit (Intel)]
>>>
>>> from django.core import management;import mysite.settings as settings;management.setup_environ(settings)
'path\\to\\mysite'
>>> from mysite.myapp.models import School
>>> School.objects.all()
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 238, in iterator
for row in self.query.results_iter():
File "C:\Python26\lib\site-packages\django\db\models\sql\query.py", line 287, in results_iter
for rows in self.execute_sql(MULTI):
File "C:\Python26\lib\site-packages\django\db\models\sql\query.py", line 2368, in execute_sql
cursor = self.connection.cursor()
File "C:\Python26\lib\site-packages\django\db\backends\__init__.py", line 81, in cursor
cursor = self._cursor()
File "C:\Python26\lib\site-packages\django\db\backends\sqlite3\base.py", line 170, in _cursor
self.connection = Database.connect(**kwargs)
OperationalError: unable to open database file
我在这里做错什么了?
发布于 2010-04-30 17:16:58
这个错误是关于没有打开数据库文件。因此,我猜想您的settings.py中的数据库路径是一个相对路径,PyDev使用与通常使用的不同的当前目录启动shell。
如果是这样的话,将DATABASE_NAME
设置更改为绝对路径,它应该可以工作。
https://stackoverflow.com/questions/2746342
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