I'm building a simple helper script for work that will copy a couple of template files in our code base to the current directory. I don't, however, have the absolute path to the directory where the templates are stored. I do have a relative path from the script but when I call the script it treats that as a path relative to the current working directory. Is there a way to specify that this relative url is from the location of the script instead?
In the file that has the script, you want to do something like this:
import os dirname = os.path.dirname(__file__) filename = os.path.join(dirname, 'relative/path/to/file/you/want')
This will give you the absolute path to the file you're looking for. Note that if you're using setuptools, you should probably use its package resources API instead.
UPDATE: I'm responding to a comment here so I can paste a code sample. :-)
Am I correct in thinking that
__file__is not always available (e.g. when you run the file directly rather than importing it)?
I'm assuming you mean the
__main__ script when you mention running the file directly. If so, that doesn't appear to be the case on my system (python 2.5.1 on OS X 10.5.7):
#foo.py import os print os.getcwd() print __file__ #in the interactive interpreter >>> import foo /Users/jason foo.py #and finally, at the shell: ~ % python foo.py /Users/jason foo.py
However, I do know that there are some quirks with
__file__ on C extensions. For example, I can do this on my Mac:
>>> import collections #note that collections is a C extension in Python 2.5 >>> collections.__file__ '/System/Library/Frameworks/Python.framework/Versions/2.5/lib/python2.5/lib- dynload/collections.so'
However, this raises an exception on my Windows machine.
os.path.realpath (sample below adds the parent directory to your path)
import sys,os sys.path.append(os.path.realpath('..'))