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.)

1
454
6/15/2015 4:06:52 PM

Accepted Answer

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
241
8/25/2012 9:57:31 PM

__file__

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

import os

os.path.realpath(__file__)

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