Asked By – Ray
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:
- In turn,
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.)
Now we will see solution for issue: How do I get the path and name of the file that is currently executing?
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