How do I get the path and name of the file that is currently executing?

Question

I have scripts calling other script files but I need to get the filepath of the file that is currently running within the process.

For example, let\'s say I have th

Answer 1

p1.py:

execfile("p2.py")

p2.py:

import inspect, os
print (inspect.getfile(inspect.currentframe()) # script filename (usually with path)
print (os.path.dirname(os.path.abspath(inspect.getfile(inspect.currentframe())))) # script directory