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

I think this is cleaner:

import inspect
print inspect.stack()[0][1]

and gets the same information as:

print inspect.getfile(inspect.currentframe())

Where [0] is the current frame in the stack (top of stack) and [1] is for the file name, increase to go backwards in the stack i.e.

print inspect.stack()[1][1]

would be the file name of the script that called the current frame. Also, using [-1] will get you to the bottom of the stack, the original calling script.

Answer 2

To get directory of executing script

 print os.path.dirname( inspect.getfile(inspect.currentframe()))

Answer 3

if you want just the filename without ./ or .py you can try this

filename = testscript.py
file_name = __file__[2:-3]

file_name will print testscript you can generate whatever you want by changing the index inside []

Answer 4

import os
print os.path.basename(__file__)

this will give us the filename only. i.e. if abspath of file is c:\abcd\abc.py then 2nd line will print abc.py

Answer 5

I think it's just __file__ Sounds like you may also want to checkout the inspect module.

Answer 6

To keep the migration consistency across platforms (macOS/Windows/Linux), try:

path = r'%s' % os.getcwd().replace('\\','/')