How can I print a Python file's docstring when executing it?

Question

I have a Python script with a docstring. When the parsing of the command-line arguments does not succeed, I want to print the docstring for the user\'s information.

Answer 1

Here is an alternative that does not hardcode the script's filename, but instead uses sys.argv[0] to print it. Using %(scriptName)s instead of %s improves readability of the code.

#!/usr/bin/env python
"""
Usage: %(scriptName)s

This describes the script.
"""

import sys
if len(sys.argv) < 2:
   print __doc__ % {'scriptName' : sys.argv[0].split("/")[-1]}
   sys.exit(0)

Answer 2

Argument parsing should always be done with argparse.

You can display the __doc__ string by passing it to the description parameter of Argparse:

#!/usr/bin/env python
"""
This describes the script.
"""


if __name__ == '__main__':
    from argparse import ArgumentParser
    parser = ArgumentParser(description=__doc__)
    # Add your arguments here
    parser.add_argument("-f", "--file", dest="myFilenameVariable",
                        required=True,
                        help="write report to FILE", metavar="FILE")
    args = parser.parse_args()
    print(args.myFilenameVariable)

If you call this mysuperscript.py and execute it you get:

$ ./mysuperscript.py --help
usage: mysuperscript.py [-h] -f FILE

This describes the script.

optional arguments:
  -h, --help            show this help message and exit
  -f FILE, --file FILE  write report to FILE

Answer 3

The docstring is stored in the module's __doc__ global.

print(__doc__)

By the way, this goes for any module: import sys; print(sys.__doc__). Docstrings of functions and classes are also in their __doc__ attribute.

Answer 4

This will print the __doc__ string when --help is the only argument.

if __name__=='__main__':
 if len(sys.argv)==2 and sys.argv[1]=='--help':
    print(__doc__)

Works for both:

• ./yourscriptname.py --help
• python3 yourscriptname.py --help