I have a script "run.py" that must print "Hello", launch another script "run2.py", and then terminate (do not wait for run2.py to return). run2.py is not in the local directory and is only required to print "Hello again".
How can I do this?
# run_path = "C:/Program Files (x86)/xxx/run.py"
# run2_path = "//network_share/folder/run2.py"
**run.py**
import os
print("Hello")
# What do I do here?
# os.execl("//network_share/folder/run2.py")
exit()
**run2.py**
print("Hello again")
This seems to work for a script I have in the same folder I'm running this one in.
This should verify that the first script finishes and doesn't linger while the second script runs in its own process. It is possible on some systems, due to their configuration, the child process will terminate when the parent does. But not in this case...
I put more time into this post to add code that shows how to check if the parent process is still running. This would be a good way for the child to ensure it's exited. Also shows how to pass parameters to the child process.
# launch.py
import subprocess as sp
import os
if __name__ == '__main__':
sp.Popen(['ps']) # Print out runniing processes.
print("launch.py's process id is %s." % os.getpid())
# Give child process this one's process ID in the parameters.
sp.Popen(['python3', 'runinproc.py', str(os.getpid())])
# ^^^ This line above anwers the main question of how to kick off a
# child Python script.
print("exiting launch.py")
Other script.
# runinproc.py
import time
import subprocess as sp
import sys
import os
def is_launcher_running():
try:
# This only checks the status of the process. It doesn't
# kill it, or otherwise affect it.
os.kill(int(sys.argv[1]), 0)
except OSError:
return False
else:
return True
if __name__ == '__main__':
print("runinproc.py was launched by process ID %s" % sys.argv[1])
for i in range(100):
if is_launcher_running():
# Is launch.py still running?
print("[[ launch.py is still running... ]]")
sp.Popen(['ps']) # Print out the running processes.
print("going to sleep for 2 seconds...")
time.sleep(2)
Bash output:
Todds-iMac:pyexperiments todd$ python3 launch.py
launch.py process id is 40975.
exiting launch.py
Todds-iMac:pyexperiments todd$ runinproc.py was launched by process ID 40975
going to sleep for 2 seconds...
PID TTY TIME CMD
PID TTY TIME CMD
40866 ttys000 0:00.09 -bash
40866 ttys000 0:00.09 -bash
40977 ttys000 0:00.04 /Library/Frameworks/Python.framework/Versions/3.8/Resources/Python.app/C
40977 ttys000 0:00.04 /Library/Frameworks/Python.framework/Versions/3.8/Resources/Python.app/C
going to sleep for 2 seconds...
PID TTY TIME CMD
40866 ttys000 0:00.09 -bash
40977 ttys000 0:00.04 /Library/Frameworks/Python.framework/Versions/3.8/Resources/Python.app/C
going to sleep for 2 seconds...
PID TTY TIME CMD
40866 ttys000 0:00.09 -bash
40977 ttys000 0:00.04 /Library/Frameworks/Python.framework/Versions/3.8/Resources/Python.app/C
going to sleep for 2 seconds...
Note that the first call to the shell, ps from launch.py is executed after launch.py exited. That's why it doesn't show up in the printed process list.
subprocess is your friend, but if you need to not wait, check out the P_NOWAIT--replacing example code in https://docs.python.org/3/library/subprocess.html
EG: pid = Popen(["/bin/mycmd", "myarg"]).pid
I don't think .communicate is what you need this time around - isn't it more for waiting?
The cleanest way to do this (since both scripts are written in pure Python) is to import the other script as a module and execute its content, placed within a function:
run.py
import os
import sys
sys.path.append("//network_share/folder/")
import run2
print("Hello")
run2.main()
exit()
run2.py
def main():
print("Hello again")
sys.path tells Python interpreter where to look for modules.
subprocessis a good python package that should help you to do that: docs.python.org/3/library/subprocess.html You should have a look at thePopen.communicatemethod.runpymodule, or read the "run2.py" and useexecfunction