Python run another script from a python script

Question

def run_cmd(self, cmd):
    print cmd 

    status = os.system(cmd)
    print 'Command ran with Status: %s' % status

    if status:
        s = 'Command Failed: Status: %s for %s' % (status, cmd)
        print s
        sys.exit(status)
    else:
        s = 'Command Success: %s' % cmd 

    return status

I am using this function to run a python script from another script.

Ex:

command_to_exec = 'python script_b.py'
run_cmd(command_to_exec)

Now if script_b.py fails script_a.py exits with the status.

That's okay.

Case 2 :

command_to_exec = 'python script_a.py --options'
rum_cmd(command_to_exec)

This case is script_a.py running 'python script_a.py' from itself. In this case, if newly spawned script_a.py fails, the outer script_a.py is still a success, because it wasn't able to catch the failure of inner script_a.py (because of sys.exit(status))

Now, how do I handle this situation? (If inner script_a fails, outer script_a should exit)

Answer 1

You can just run your other script as a module. Let's say you have the folder:
dir
|---main.py
|---imported.py
And you want to run the second from the first. Just use import , like that:

import imported #the file name without the extension

It will run that file, and you will be able to use it's vars and functions in your main script.