获取导入文件的__file__

Question

Say there is a file tree of:

sample_module
  |___ file1.py
  |___ file2.py

How can I get file2 's __file__ , from a function in file1 ?

file1.py :

def get_file_path():
    return __file__

file2.py

from sample_module.file1 import get_file_path
print(get_file_path())

# C:/.../sample_module/file1.py

How can I make it give me C:/.../file2.py ?

I understand that the function get_file_path is saving the __file__ into its namespace, and is not generating a new __file__ . Is there a way to make __file__ callable, or get the __file__ at runtime?

---

Why I need it?

I have a library, with a logging module, that generates a logger.

I want the location of that logger to be eg 2 folders above file importing generate_logger . I'd rather not the user have to specify __file__ each time they want to use it. Is there a better way?

ie in mylibrary/logs.py

def generate_logger(logname=__name__, level=logging.DEBUG, path=''):
    """Creates a logger instance. Set level of file handler.
    """
    logger = logging.getLogger(logname)
    if path == '':
        path = os.path.abspath(os.path.join(os.path.dirname(__file__), '../../logs'))
    fh = logging.FileHandler(f'{path}/{logname}.log')
    ## rest of logging code.
    return logger

-

 project
  |___ project
  |      |__ log
  |            |__ loads_generate_logger (code that does the import)
  |_______logs
            |_logname.log (creates a log folder 2 folders above where it was imported).       

in loads_generate_logger

 from mylibrary import generate_logger()
 logger = generate_logger()
 # logger path = C:/.../venv/lib/site-packages/mylib-0.10/logs

Answer 1

you can try this:

import inspect

current_frame = inspect.currentframe()

current_frame.f_back.f_globals['__file__']

this will get the __file__ global from the function that called your function