Converting a windows path to a path using environment variables

Question

Is there a Python library that takes a windows path and replaces expanded environment variables with environment variables?

For example:

C:\Users\Username\Documents\Text.txt -> %USERPROFILE%\Documents\Text.txt
C:\Windows\System32\cmd.exe -> %WINDIR%\System32\cmd.exe
C:\Program Files\Program\Program.exe -> %PROGRAMFILES%\Program\Program.exe

The best way to explain would be to get functionality that does the opposite of

os.path.expandvars('some path with environment variables')

Support for different languages would also be a requirement.

C:\Archivos de programa\Progra\Program.exe -> %PROGRAMFILES%\Program\Program.exe

I don't think that is possible, or at least it is ambiguous. There is nothing to say that there is a one-to-one mapping of directories to environment variables. In other words I could have both %FOO% and %BAR% that both expand to C:\Windows, in that case which should I use? — Cory Kramer
– Cory Kramer, Commented Jul 25, 2017 at 12:43
There is a winshell module, specifically its Special Folders part that grabs some of the special folders. For the rest you'll have to use Win32 API. Then it's just a matter of checking if your path intersects with the special folder and replacing it with its canonical environment variable name - you can use the os.path facilities for that. — zwer
– zwer, Commented Jul 25, 2017 at 12:46
@CoryKramer it only needs to map to known default windows environment variables, not any new ones. — HJ Blom
– HJ Blom, Commented Jul 25, 2017 at 12:52

Jean-François Fabre · Accepted Answer · 2017-07-25 13:06:01Z

1

This is a non-trivial problem since more than environment variables may match parts of your string (ex: PROCESSOR_LEVEL is usually a single digit, so you should avoid it). To ensure the best efficiency, I would:

sort the existing environment variable l

like this:

import os

my_string = os.path.normpath(r"D:\Users\JOTD\AppData\Roaming\Adobe\Flash Player")


for k,v in sorted(os.environ.items(),key=lambda x:len(x[1]),reverse=True):
    my_new_string = my_string.replace(v+os.sep,"%{}%{}".format(k,os.sep))
    if my_string != my_new_string:
        break
    my_string = my_new_string

print(my_new_string)

answered Jul 25, 2017 at 13:06

Jean-François Fabre♦

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