Start and stopping external program?

Question

I want to start and stop an external program in Python3, on Linux. This program is an executable, called smip-tun, located in the same folder as the source code and I'd like to be able to execute and control it from there with relative names.

Unfortunately, this does not work:

smip_tun = 0

def start_smip_tun(self):
    smip_tun =  subprocess.Popen(['smip-tun',DEFAULT_SERIAL_PORT])

def end_smip_tun(self):
    smip_tun.terminate()
    print("end")

But says it cannot find smip-tun. Efforts to specify the relative directory have failed. I was trying with cwd but cannot figure it out.
Any advice?

Edit:

Made Smip-tun global but the issue persists.

New code:

smip_tun = 0

def start_smip_tun(self):
    global smip_tun
    smip_tun =  subprocess.Popen(['smip-tun',DEFAULT_SERIAL_PORT])

def end_smip_tun(self):
    global smip_tun
    smip_tun.terminate()
    print("end")

Error message:

  File "/home/sven/git/cerberos_manager/iot_network_api.py", line 40, in start_smip_tun
    smip_tun = subprocess.Popen(['smip-tun',DEFAULT_SERIAL_PORT])
  File "/usr/lib/python3.5/subprocess.py", line 947, in __init__
    restore_signals, start_new_session)
  File "/usr/lib/python3.5/subprocess.py", line 1551, in _execute_child
    raise child_exception_type(errno_num, err_msg)
FileNotFoundError: [Errno 2] No such file or directory: 'smip-tun'

You need to declare `smip_tun` global, see [Using global variables in a function other than the one that created them](http://stackoverflow.com/q/423379/45249) — mouviciel, Nov 28 '16 at 12:47
More specifically, add a `global smip_tun` statement to the beginning of the `start_smip_tun()` function. — martineau, Nov 28 '16 at 13:08
You had two issues. The `global` advice solved the one you didn't see because of the second one. Now, you have to specify the path of `smip-tun` or include that path in your `PATH` environment variable (e.g., `export PATH=.:$PATH`). — mouviciel, Nov 29 '16 at 10:23
Yes indeed, but my main question was on how to do that relative to the directory the script itself is executing in. Smip-tun is located in the same directory as the .py file. — , Nov 29 '16 at 12:10
I am a bit slow. The mistake is that I need to specify "./smip-tun" to execute, not smip-tun — , Nov 29 '16 at 12:28

score 1 · Accepted Answer · edited Nov 29 '16 at 14:19

1

import os

os.chdir('.') # or change the . for the full path directory.

smip_tun =  subprocess.Popen(['smip_tun',DEFAULT_SERIAL_PORT])

Also make sure your app do not have any extension, like smip-tun.py and smip-tun.sh

edited Nov 29 '16 at 14:19

Wondercricket

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answered Nov 29 '16 at 13:51

Rafael Rodrigo De Souza

160
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Start and stopping external program?

1 Answers1