我建立了一个新项目,并用简单的模型填充它。(基本上我是在跟着tut。)
当我python manage.py shell
在命令行上运行时,它工作正常:
>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 mysite 项目 >> Django >> Shell with Django environment
这是 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
我在这里做错了什么?