Find the realpath for a symlinked .pth file in itself with python code.

[hongyi]

Hi,

I symlinked a pth file to the site-packages directory of my virtualenv. Then I want to obtain the realpath of this pth file from within it with python code.

But it seems the python code used in this file must only be one-line python code and it doesn't know the meaning of 'os.path.realpath(__file__)'

Any hints for this problem?

Regards

[Larz60+]

import os

os.chdir(os.path.abspath(os.path.dirname(__file__))) # Make sure starting dir is OK

print(os.path.realpath('CreateDict.py'))

[hongyi]

This doesn't solve my problem, I want to obtain the directory from where the pth file is symlinked to the site-packages directory, instead of the site-packages directory itself. Say, if I symlinked the the file with

ln -s ~/test.pth  /path/to/site-packages/ 

And there are the following the lines in this pth file:

import os; os.chdir(os.path.abspath(os.path.dirname(__file__))); print(os.path.realpath('CreateDict.py')) 

This will print the following twice:

$ python
/home/werner/.pyenv/versions/3.8.1/lib/python3.8/CreateDict.py
/home/werner/.pyenv/versions/3.8.1/lib/python3.8/CreateDict.py
Python 3.8.1 (default, Mar 11 2020, 09:08:36) 
[GCC 9.2.1 20191008] on linux
Type "help", "copyright", "credits" or "license" for more information.
>>> 

But I want the obtain the directory of $HOME.

Regards

[anbu23]

from os.path import expanduser
home = expanduser("~")

[hongyi]

@anbu23 I just use HOME as a simple example, in fact the pth file is not in that directory.

$ ln -s /path/to/test.pth /path/to/site-packages/

In this general case, I want to obtain the source path of the pth file use python code from within it.

Regards

[anbu23]

print(os.path.abspath('test.pth'))

[hongyi]

No, it cannot do the trick.

[Larz60+]

You cannot use __file__. It will always return the current script name, as will sys.argv[0]
you must use actual file name:

import os
import sys


def show_symlink():
    # Set current path to that of script
    os.chdir(os.path.abspath(os.path.dirname(__file__)))

    # get address of self
    print(f"value of __file__: {__file__}")
    print(f"value of sys.argv[0]: {sys.argv[0]}")

    print(f"Current working directory: {os.getcwd()}")

    print(f"using realpath: {os.path.realpath('CreateDict.py')}")
    print(f"using readlink: {os.readlink('CreateDict.py')}")

show_symlink()

will give:

Output:
value of __file__: .../projects/T-Z/T/Trailmapper/Florida2020/src/TestArea/ShowSympath.py
value of sys.argv[0]: .../projects/T-Z/T/Trailmapper/Florida2020/src/TestArea/ShowSympath.py
Current working directory: .../projects/T-Z/T/Trailmapper/Florida2020/src/TestArea
using realpath: .../projects/T-Z/T/Trailmapper/src/CreateDict.py
using readlink: ../CreateDict.py

As you can see, print(f"using realpath: {os.path.realpath('CreateDict.py')}") gives the full directory of what the symlink points to
As does print(f"using readlink: {os.readlink('CreateDict.py')}") albeit that this is a relative path (exactly as originally defined with the ln command)

CreateDict.py resides one directory above TestArea (in the src Directiry), the symlink is in TestArea

This works properly.