launch an exe with parameter from python

Question

I'm trying to call an application and pass a text file as a parameter. The text file has instructions. I tried passing a parameter when opening notepad but couldn't get it to work. Here is the code I'm trying:

import os

os.startfile(r"C:\Windows\notepad.exe c:\cobra\advancecalendar.txt")

If I don't include the path to the second file it launches notepad, but with the file name included i get the following error:

RESTART: C:/Users/timbo/AppData/Local/Programs/Python/Python36-32/Open exe module.py 
Traceback (most recent call last):
  File "C:/Users/timbo/AppData/Local/Programs/Python/Python36-32/Open exe module.py", line 3, in <module>
    os.startfile(r"C:\Windows\notepad.exe c:\cobra\advancecalendar.txt")
FileNotFoundError: [WinError 2] The system cannot find the file specified: 'C:\\Windows\\notepad.exe c:\\cobra\\advancecalendar.txt'

How can I pass a parameter to the startfile so it opens a specific file? when I tried putting both in quotes it still got an error.

Possible duplicate https://stackoverflow.com/questions/15928956/how-to-run-an-exe-file-with-the-arguments-using-python — Nathan, Mar 14 '18 at 18:12
Try searching a bit more, I think this gets asked pretty often. — Bemmu, Mar 14 '18 at 18:13
Use `os.startfile()` just with the file, without the application. — Klaus D., Mar 14 '18 at 18:13
Thanks. I didn’t see one with parameters but was probably to narrow in my search will dig in again — Tim Boatwright, Mar 16 '18 at 12:53

score 2 · Answer 1 · answered Mar 14 '18 at 18:15

2

Use the subprocess module.

import subprocess
subprocess.call([r"C:\Windows\notepad.exe", "c:\cobra\advancecalendar.txt"])

Tested in python2.7

answered Mar 14 '18 at 18:15

Rakesh

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Thanks’ that should solve another problem I was having too. – Tim Boatwright Mar 16 '18 at 12:52
You are welcome. Please accept ans if it solved your problem. Thanks – Rakesh Mar 16 '18 at 12:54

launch an exe with parameter from python

1 Answers1