Execution error when Passing arguments to a python script using os.system. The script takes sys.argv arguments

Question

I have tried to execute a simple python command from cmd like C:\Users> stat.py < swagger.yaml > output.html, which executes stat.py by taking swagger.yaml as input argument and generates output.html file and it worked fine in cmd. But now i want to execute my stat.py file through another python file demo.py by passing the values swagger.yaml and output.html as sys.argv[0] and sys.argv[1] inside demo.py.

my command from cmd C:\Users> demo.py swagger.yaml output.html and my demo.py file is as follows..

 # my demo.py file ....

import os
import sys

os.system('stat.py < sys.argv[1] > sys.argv[2]')

error - the system can not find the file specified.
Why i am getting this error and please any help to resolve it ..

ShadowRanger · Accepted Answer · 2019-08-17 02:54:17Z

Inside a normal string, no variable interpretation is applied. So you literally asked to read from a file named sys.argv[1] (possibly sys.argv1 if the file exists, thanks to shell globbing), and write to a file named sys.argv[2].

If you want to use the values sys.argv in your script, you need to format them into the string, e.g. with f-strings (modern Python 3.6 or so only):

os.system(f'stat.py < {sys.argv[1]} > {sys.argv[2]}')  # Note f at beginning of literal

or on older Python 2.7, with str.format:

os.system('stat.py < {} > {}'.format(sys.argv[1], sys.argv[2]))

Note that however you slice it, this is dangerous; os.system is launching this in a shell, and arguments that contain shell metacharacters will be interpreted as such. It can't do anything the user didn't already have permission to do, but small mistakes by the user could dramatically change the behavior of the program. If you want to do this properly/safely, use subprocess, open the files yourself, and pass them in explicitly as stdin/stdout:

with open(sys.argv[1], 'rb') as infile, open(sys.argv[2], 'wb') as outfile:
    subprocess.run(['stat.py'], stdin=infile, stdout=outfile)

This ensures the files can be opened in the first place before launching the process, doesn't allow the shell to interpret anything, and avoids the (minor) expense of launching a shell at all. It's also going to give you more useful errors if opening the files fails.

Collectives™ on Stack Overflow

Execution error when Passing arguments to a python script using os.system. The script takes sys.argv arguments

1 Answer 1

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

Comments

Your Answer

Sign up or log in

Post as a guest

Related