CODEWITHSUNDEEP
python
turd
turd

Reputation: 27

How can I launch a process in Python on a raspberry pi 3 raspbian?

I would like to launch a process from a Python script on a raspberry pi 3 raspbian.

On Windows, the following code would work:

import os
def openFile():
    try:
        os.startfile("/home/pi/Desktop")
    except:
        print("invalid path")

But here I get back invalid path.

Can you tell me how to fix this and also how to open applications?

I remember that in windows: There's Notepad.exe in a path and I can open it but what should I say for Linux? I mean what is the .exe in Linux and can I open it?

Upvotes: 1

Views: 682

Answers (2)

Kolay.Ne
Kolay.Ne

Reputation: 1398

As far as I understand, what you're trying to do is called not open but execute. So you could find more info about that by searching for "python execute file on Linux". Despite this, you are trying to execute a directory, not a file. So, here is an example of what I'd do:

import subprocess


subprocess.call(['/bin/ls', '-l'])

This will call the executable file ls located at the /bin folder and provide it with one argument: -l. It will list files in your current directory (however, remember that you shouldn't use ls for that purpose, this is just an example. If you want to list files in a directory, there are special functions for that in Python).

Speaking about file extensions, the executable file on Linux (similar to Windows' .exe file) is called an ELF file, which does not have a canonical extension. In fact, Linux in general cares about extensions much less, than Windows. If you want to know more about which else files can be executed on Linux, execute permissions etc, please, search for info on the internet and/or ask a question at https://superuser.com)

Upvotes: 2

Adam Oudad
Adam Oudad

Reputation: 347

os.startfile is only available for Windows. You should use instead the subprocess library. Try this platform-independant solution from @user4815162342 which I adapted

import os, sys, subprocess

def open_file(filename):
    if sys.platform == "win32":
        os.startfile(filename)
    else:
        opener = "open" if sys.platform == "darwin" else "xdg-open"
        subprocess.run([opener, filename])

If your file is just a bash script, you can replace the subprocess.run line by

subprocess.run(["bash", filename])

Upvotes: 2

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