Re: Developing a Package with Sub Packages
En Sun, 17 Feb 2008 22:34:27 -0200, Josh English [EMAIL PROTECTED] escribi�: Here's what I think is happening: IMS/__init__.py uses os.getcwd() to establish the path to the data folder and the files inside of it. When I run StoryCreator, os.getcwd() returns the story folder. If I'm right, how can I get the IMS/__init__.py module to use relative paths to its own module, and not the current working directory the os module provides? Use the module's own __file__ attribute: my_path = os.path.abspath(os.path.dirname(__file__)) -- Gabriel Genellina -- http://mail.python.org/mailman/listinfo/python-list
Re: Developing a Package with Sub Packages
When testing the package in idle, this results in C:\Python25\Lib\idlelib instead of the file. The Data folder is created in this folder now. On 2/18/08, Gabriel Genellina [EMAIL PROTECTED] wrote: En Sun, 17 Feb 2008 22:34:27 -0200, Josh English [EMAIL PROTECTED] escribi�: Here's what I think is happening: IMS/__init__.py uses os.getcwd() to establish the path to the data folder and the files inside of it. When I run StoryCreator, os.getcwd() returns the story folder. If I'm right, how can I get the IMS/__init__.py module to use relative paths to its own module, and not the current working directory the os module provides? Use the module's own __file__ attribute: my_path = os.path.abspath(os.path.dirname(__file__)) -- Gabriel Genellina -- http://mail.python.org/mailman/listinfo/python-list -- Josh English [EMAIL PROTECTED] http://joshenglish.livejournal.com -- http://mail.python.org/mailman/listinfo/python-list
Re: Developing a Package with Sub Packages
En Mon, 18 Feb 2008 16:23:28 -0200, Josh English [EMAIL PROTECTED] escribi�: When testing the package in idle, this results in C:\Python25\Lib\idlelib instead of the file. The Data folder is created in this folder now. Works for me: main.py: from testpkg import a testpkg directory (a subdirectory somewhere in sys.path): testpkg\__init__.py: (empty file) testpkg\a.py: import os print __name__, __file__, os.path.abspath(os.path.dirname(__file__)) In IDLE: File - Open, main.py. F5 (Run Module). Output: testpkg.a C:\APPS\PYTHON25\lib\site-packages\testpkg\a.pyc C:\APPS\PYTHON25\lib\site-packages\testpkg -- Gabriel Genellina -- http://mail.python.org/mailman/listinfo/python-list
Developing a Package with Sub Packages
I have created a group of scripts to manage an XML-based database. I'd like to make it into a proper package that will let me keep track of the code. I have a lot of files that are similar in name and they just get crowded in one folder. Here's a sample of the file structure: IMS/ IMS/__init__.py IMS/Config.py IMS/imsdefaults.cfg IMS/local.txt IMS/Data/ IMS/Data/stories.xml IMS/Story/ IMS/Story/__init__.py IMS/Story/StoryCreator.py When the IMS/__init__.py file is loaded, it creates the IMS/Data/ folder and if there are no xml files, it creates them and fills them in with some default values. The IMS/Config.py has a subclass of the ConfigParser, and reads the imsdefaults.cfg and local.txt files. When I run StoryCreator, buried in it's own package (that imports IMS) the data folder is created inside the Story folder, and I get an error message stating the ConfigParser object could not find the imsdefaults.cfg file. Here's what I think is happening: IMS/__init__.py uses os.getcwd() to establish the path to the data folder and the files inside of it. When I run StoryCreator, os.getcwd() returns the story folder. If I'm right, how can I get the IMS/__init__.py module to use relative paths to its own module, and not the current working directory the os module provides? This could also solve the problem with Config.py, I think. Thanks Josh English http://joshenglish.livejournal.com -- http://mail.python.org/mailman/listinfo/python-list
