Is it possible to obtain the path of the script us

Question

I would like to know how to get the path where the script is stored with argparse, if possible, because if I run the script from another path (I have the path of the script in the %PATH% variable) it uses by default the relative path.

I know that I can obtain it using:

import sys sys.argv[0]

but I would like to know if it is possible to acess it directly from the argparse module.

Thanks Edit: I have my reply and I am satisfied. To explain better the question: I have a script called mdv.py that I use to transform markdown files into html. I would like to call it from any location in my computer. The script is in: c:\Python27\markdown

in this path there are other files and a folder templates that I use to generate my HTML (a default stylesheet and files for header, body and footer).

These files are in: C:\Python\27\markdown\markdown\templates

When I call the script from a non standard path, for example c:\dropbox\public it looks in c:\dropbox\public\templates for these files and not in c:\python27\markdown\templates where they are saved.

Ihope to have better explained. Sorry I'm not a native english speaker.

Answer 1

I think you are looking for the prog parameter; you can interpolate sys.argv[0] into your help strings with %(prog)s.

The value for prog can be set when creating the ArgumentParser() instance; it is the first parameter:

parser = argparse.ArgumentParser('some_other_name')

and can be retrieved with the .prog attribute:

print(parser.prog)  # prints "some_other_name"

However, argparsecalls os.path.basename() on this name, and does not store the directory of the program anywhere.