How to print the filename of the current execution file in python?

Asked Apr 24 '13 at 12:50

Active Apr 24 '13 at 12:50

Viewed 68 times

I am currently running a script which while execution goes to multiple other python files. My aim is to print the filename of the currently executing file in python ?

I tried using:

print __file__

But the following error was thrown:

Traceback (most recent call last):
  File "C:\Users\qksr\Desktop\work\parsing text\testcase.py", line 32, in <module>
    print __file__
NameError: name '__file__' is not defined

So what statement should i code ?

asked Apr 24 '13 at 12:50

Kaushik

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It works for me. What is your OS? – xylon97 Apr 24 '13 at 12:52
@xylon97 mine is windows – Kaushik Apr 24 '13 at 12:53
I believe __file__ is undefined when __name__ == '__main__'. – Pace Apr 24 '13 at 13:12
How is `testcase.py` being executed? – martineau Apr 24 '13 at 13:56
What does a `print __name__` display if put before the line? – martineau Apr 24 '13 at 14:06
@martineau its displays: '__main__' – Kaushik Apr 25 '13 at 09:47
Try `print os.path.abspath(inspect.getsourcefile(lambda _: None))` following an `import os, inspect` statement. – martineau Apr 25 '13 at 16:19

How to print the filename of the current execution file in python?

0 Answers0