I have 4 python script files.

• script_1.py
• script_2.py
• script_3.py
• script_4.py

script_1.py, script_2.py and script_3.py calls script_4.py

Execute files always change.

Like, python script_1.py / python script_2.py / python script_3.py

I want to get script_1.py, script_2.py, and script_3.py file name in script_4.py python script.

How can I get current executing file name without having to pass that information as arguments?

I tried __name__ and inspect.getfile(inspect.currentframe()) but no luck.

Both of codes are returns script_4.py.

CodePudding user response：

When called, sys.argv holds the command line arguments. The first element is the name of the script that was called (including path). To get the name of the script that was originally called you can do, in any of your files:

import os
import sys

called_script_with_path = sys.argv[0]
called_script_without_path = os.path.basename(sys.argv[0])

CodePudding user response：

You can use the module level attribute __file__ to get the path to the current module.

>>> import os
>>> os.path.basename(__file__)
test.py

If you want to determine this for another file, you can import the script and still query the value:

>>> import os
>>> import script1
>>> os.path.basename(script1.__file__)
script1.py