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