Getting the correct file path on case-insensitive systems

Python Help
1

On such systems like Windows and macOS, is there a way to get the path as stored on disk without resolving symbolic links? Context is that within a virtual environment the paths that sysconfig returns are in all caps for some reason:

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cc @barneygale the current path expert :slightly_smiling_face:

2

What do listdir and other such functions return? Do they return the actual case?

If they do, then that’s one way.

3

It does return the actual case but that means I would have to do that for every path component.

4

A quick look reveals os.path.realpath:

>>> os.path.realpath('C:/PYTHON312')
'C:\\Python312'
5

Yes that definitely works but I’m trying to find a way without resolving symbolic links like I mentioned.

6

Is this a windows only problem? If so symlinks are not an issue?
So you could call os.path.realpath only is on Windows?

7

This also affects macOS. I don’t want to resolve because it’s possible that an input path must be a symbolic link, I just want the actual name.

8

It feels like this must be happening in get_paths, maybe environment variables are being changed in the venv? Or it thinks the OS is different for some reason…

It doesn’t happen in my env but i’m not using hatch.

9

On macOS with python 3.11 I cannot reproduce this problem:

% python3.11
Python 3.11.5 (v3.11.5:cce6ba91b3, Aug 24 2023, 10:50:31) [Clang 13.0.0 (clang-1300.0.29.30)] on darwin
Type "help", "copyright", "credits" or "license" for more information.
>>> import sysconfig
>>> print(sysconfig.get_path('include'))
/Library/Frameworks/Python.framework/Versions/3.11/include/python3.11

Also I cannot reproduce on windows 11.

py -3.11
Python 3.11.6 (tags/v3.11.6:8b6ee5b, Oct  2 2023, 14:57:12) [MSC v.1935 64 bit (AMD64)] on win32
Type "help", "copyright", "credits" or "license" for more information.
>>> import sysconfig
>>> print(sysconfig.get_path('include'))
C:\Program Files\Python311\Include
>>>

Also works when I create a venv.

Something your code is doing to the value?

10

Still, I would like to know if there is a way to get the correct path name without resolving symbolic links.

11

For each element in the path you would need to os.listdir() and pick the matching entry.

1 Like
12

One way to do it:

def realcase(path):
    dirname, basename = os.path.split(path)
    if dirname == path:
        return dirname
    dirname = realcase(dirname)
    basename = os.path.normcase(basename)
    for child in os.listdir(dirname):
        if os.path.normcase(child) == basename:
            return os.path.join(dirname, child)
    raise FileNotFoundError(f'{path} not found')
1 Like