在Python中的Linux上的子目录中找到顶级目录

Question

I am creating various HTML file parts, image thumbnails etc. from within a CodeIgniter application tree using cron-scheduled Python 2.7 programs on Linux. The actual Python programs exist under the CodeIgniter tree in a subdirectory one level below the application directory as follows.

codeigniter/web-root 
    |
    application
    |   |    
    |   scripts
    |   |   |
    |   |   my-program.py
    |   | 
    |   database
    |       |
    |       database.sqlite 
    images

I want to determine the codeigniter/web-root directory from within my-program.py using methods from the os.path module. However, the absolute path to the codeigniter/web-root is different on the development and production environments so I prefer not to hardwire this path information into the Python program itself.

The current script(s) use the following construct to determine the absolute path of "codeigniter/web-root" which is two directory levels above the script itself in both environments.

#!/bin/env python2.7

import os.path

ci_root = os.path.dirname(os.path.dirname(os.path.dirname(os.path.abspath(__file__))))

Is there a cleaner way to determine the top level(ci_root) directory without using multiple os.path.dirname calls?

Answer 1

import os.path
print os.path.abspath(__file__+'/../../..')   # the directory three levels up

I was pleasantly surprised that

• abspath() managed to parse correctly, without using os.path.join()
• We didn't have to strip out the filename before we build the path

For cross platform compatibility if abspath's parsing does not work, we could use something less readable, eg

• os.path.sep instead of '/'
• os.path.join()

Answer 2

Find the real path based on the relative path of "up three directories," and extract the last part of that real path.

[Edit: In response to your comment I have revised this. In my opinion it is getting complex enough that your chain of dirname 's is better. At least it is easy to understand what it happening.]

from os import path as p
_, ci_root = p.split(p.abspath(p.join(__file__, '../../..')))