Using Python's os.path, how do I go up one directory?

Question

I recently upgrade Django from v1.3.1 to v1.4.

In my old settings.py I have

TEMPLATE_DIRS = (     os.path.join(os.path.dirname( __file__ ), 'templates').replace('\\', '/'),     # Put strings here, like "/home/html/django_templates" or "C:/www/django/templates".     # Always use forward slashes, even on Windows.     # Don't forget to use absolute paths, not relative paths. ) 

This will point to /Users/hobbes3/Sites/mysite/templates, but because Django v1.4 moved the project folder to the same level as the app folders, my settings.py file is now in /Users/hobbes3/Sites/mysite/mysite/ instead of /Users/hobbes3/Sites/mysite/.

So actually my question is now twofold:

1. How do I use os.path to look at a directory one level above from __file__. In other words, I want /Users/hobbes3/Sites/mysite/mysite/settings.py to find /Users/hobbes3/Sites/mysite/templates using relative paths.
2. Should I be keeping the template folder (which has cross-app templates, like admin, registration, etc.) at the project /User/hobbes3/Sites/mysite level or at /User/hobbes3/Sites/mysite/mysite?

Answer 1

os.path.abspath(os.path.join(os.path.dirname( __file__ ), '..', 'templates')) 

As far as where the templates folder should go, I don't know since Django 1.4 just came out and I haven't looked at it yet. You should probably ask another question on SE to solve that issue.

You can also use normpath to clean up the path, rather than abspath. However, in this situation, Django expects an absolute path rather than a relative path.

For cross platform compatability, use os.pardir instead of '..'.

Answer 2

To get the folder of a file just use:

os.path.dirname(path)  

To get a folder up just use os.path.dirname again

os.path.dirname(os.path.dirname(path)) 

You might want to check if __file__ is a symlink:

if os.path.islink(__file__): path = os.readlink (__file__)