how to run an exe file with the arguments using python

Question

Suppose I have a file RegressionSystem.exe. I want to execute this executable with a -config argument. The commandline should be like:

RegressionSystem.exe -config filename

I have tried like:

regression_exe_path = os.path.join(get_path_for_regression,'Debug','RegressionSystem.exe')
config = os.path.join(get_path_for_regression,'config.ini')
subprocess.Popen(args=[regression_exe_path,'-config', config])

but it didn't work.

Answer 1

You can also use subprocess.call() if you want. For example,

import subprocess FNULL = open(os.devnull, 'w')    #use this if you want to suppress output to stdout from the subprocess filename = "my_file.dat" args = "RegressionSystem.exe -config " + filename subprocess.call(args, stdout=FNULL, stderr=FNULL, shell=False) 

The difference between call and Popen is basically that call is blocking while Popen is not, with Popen providing more general functionality. Usually call is fine for most purposes, it is essentially a convenient form of Popen. You can read more at this question.

Answer 2

For anyone else finding this you can now use subprocess.run(). Here is an example:

import subprocess
subprocess.run(["RegressionSystem.exe", "-config filename"])

The arguments can also be sent as a string instead, but you'll need to set shell=True. The official documentation can be found here.