l have two scripts:
main.py
import package.py
package.py
import os
print(os.path.basename(_file_))
My expected output is main.py, but I'm getting package.py.
So how can l get the running script's filename in a package script?
extra description:
The truth is, I have a decorator function in package.py. It will generate a file at current path and named as the file's name who called it.
You should be able to use arguments to work it out.
from sys import argv
print(argv[0])
The first argument listed will be the command used to execute your script.
So if you're running ./main.py then that's what you'll get.
If you're running it via python (such as python main.py) then (at least according to my testing) you'll get the full path. You can use the tools in os.path to pluck out just the filename if required.
Try this
print(os.path.basename(__name__))
package.pytry:
main.py:
import package.py
package.py:
import sys
print(sys.argv[0])
If you are trying to get the entire path to the importing script, do this:
import os
import sys
# get working directory
dir = os.getcwd()
# get script path as passed to python
script = sys.argv[0]
# if path is not absolute, then join with current dir
importing_script = os.path.join(dir,script) if not os.path.isabs(script) else script
# Your importing script is...
print(importing_script)
package.pyyou could just hard-code it (not saying there is no better solution).