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 three files. Using execfile:

• script_1.py calls script_2.py.
• In turn, script_2.py calls script_3.py.

How can I get the file name and path of script_3.py, from code within script_3.py, without having to pass that information as arguments from script_2.py?

(Executing os.getcwd() returns the original starting script's filepath not the current file's.)

Answer 1

__file__ 

as others have said. You may also want to use os.path.realpath to eliminate symlinks:

import os  os.path.realpath(__file__) 

Answer 2

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