Using python, how to find a different python script in a directory and then run it

user_10270861 0
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Using python, I need to find another python script file in a directory and then run it

system = input("Enter system name: ")
for filename in listdir(directory):
   if filename.find(system + "_startup") != -1 and filename.endswith(".py"):
      # import and run specific startup script

I know how to find and open a file normally, and I know how to call one python script from another script, but I don't really know how to bridge the gap here.

Each system I'm working with will have a different "startup" script which runs. I don't want to have to import every single startup file I have into my main script (there's a lot of startup scripts) only the specific one for the system of interest. Is there a way for me to achieve this without importing all the startup files?

python
3 Answers

If all you need to do is execute the script, one option is to spawn a new process:

import subprocess

subprocess.run(["python3", filename])
user_13023170

Why not use os.system to execute the command in a subshell?

os.system('py -3 ' + filepath)
user_84144320

for catch output bash 

import subprocess
from subprocess import Popen

system = input("Enter system name: ")

for filename in listdir(directory):
   if filename.find(system + "_startup") != -1 and filename.endswith(".py"):
       p = Popen(["python3",filename], stdout=subprocess.PIPE, stderr=subprocess.PIPE)
       output, errors = p.communicate()
       # catch output bash
       print(output)
       # catch erro bash
       print(errors)
user_105690270
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