从具有运行脚本的目录中读取文件，而不是从调用脚本的位置调用目录位置python

Question

suppose I have following

test/
    Main.py
test/one/
    One.py
    test1.txt
test/two/
    Two.py
    test2.txt

What I want to do is, run Main.py and call One.py and Two.py and want to read files. Since directory containing One.py has one file test1.txt, I want to read that file and directory containing Two.py has one file test2.txt

This can be done simply by

open("one/test1.txt", "r")

and open("two/test2.txt", "r") in two files

Is there any other way to read files. This is simple case I took.

I am facing this problem. For this, I have created setting object and provide required location to that directory manually and passed this setting object to all class that needs location to read file

Is there different approach?

Answer 1

If I understand your question correctly, you want to know the directory of the running python file?

You can access the special variable __file__ inside a Python program. It returns the path of the python file.

To open a named file in the running script's directory, use:

import os
f = open(os.path.join(os.path.dirname(__file__), "filename.txt"))