How can I get the name of the script calling the function in python?

Question

I know that __file__ contains the filename containing the code, but is there a way to get the name of the script/file that's calling the function?

If I have a file named filenametest_b.py:

def printFilename():
    print(__file__)

And I import the function in filenametest_a.py:

from filenametest_b import *

printFilename()

I get:

C:\Users\a150495>python filenametest_a.py
C:\Users\a150495\filenametest_b.py

Is there something I can do in the b file to print the name of the a file?

you could probably parse sys.argv[0]

mgilson
– mgilson

2012-12-20 22:23:51 +00:00
Commented Dec 20, 2012 at 22:23 — mgilson
– mgilson, Commented Dec 20, 2012 at 22:23

Martijn Pieters · Accepted Answer · 2012-12-20 22:30:00Z

5

You could print sys.argv[0] to get the script filename.

To get the filename of the caller, you need to use the sys._getframe() function to get the calling frame, then you can retrieve the filename from that:

import inspect, sys

print inspect.getsourcefile(sys, sys._getframe(1))

edited Dec 20, 2012 at 22:30

answered Dec 20, 2012 at 22:23

Martijn Pieters

1.1m326 gold badges4.2k silver badges3.4k bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

Chris Matta Over a year ago

inspect.getsourcefile(sys._getframe(1)) works fine. Thanks.

Chris Matta · Accepted Answer · 2012-12-20 22:25:39Z

0

Asked a bit too soon, I think I just solved it.

I was just looking around in the inspect module:

I added this to the filenametest_b.py file:

def printCaller():
    frame,filename,line_number,function_name,lines,index=\
        inspect.getouterframes(inspect.currentframe())[1]
    print(frame,filename,line_number,function_name,lines,index)

which gives this:

C:\Users\a150495\filenametest_b.py
(<frame object at 0x000000000231D608>, 'filenametest_a.py', 5, '<module>', ['printCaller()\n'], 0)

So this should work.

answered Dec 20, 2012 at 22:25

Chris Matta

3,4833 gold badges39 silver badges49 bronze badges

Comments

Andre Holzner · Accepted Answer · 2012-12-20 22:30:19Z

0

def printFilename():
    import traceback 
    print traceback.extract_stack()[-2][0]

answered Dec 20, 2012 at 22:30

Andre Holzner

18.8k6 gold badges59 silver badges66 bronze badges

Comments

Ahmad · Accepted Answer · 2022-11-10 16:12:32Z

0

this work for me:

    inspect.getouterframes(inspect.currentframe())[1].filename

answered Nov 10, 2022 at 16:12

Ahmad

192 bronze badges

Collectives™ on Stack Overflow

How can I get the name of the script calling the function in python?

4 Answers 4

1 Comment

Comments

Comments

Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

4 Answers 4

1 Comment

Comments

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related