Python open() requires full path

Solution 1:

From the documentation:

open(file, mode='r', buffering=-1, encoding=None, errors=None, newline=None, closefd=True, opener=None)

file is a path-like object giving the pathname (absolute or relative to the current working directory) of the file to be opened or an integer file descriptor of the file to be wrapped.

So, if the file that you want open isn't in the current folder of the running script, you can use an absolute path, or getting the working directory or/and absolute path by using:

import os
# Look to the path of your current working directory
working_directory = os.getcwd()
# Or: file_path = os.path.join(working_directory, 'my_file.py')
file_path = working_directory + 'my_file.py'

Or, you can retrieve your absolute path while running your script, using:

import os
# Look for your absolute directory path
absolute_path = os.path.dirname(os.path.abspath(__file__))
# Or: file_path = os.path.join(absolute_path, 'folder', 'my_file.py')
file_path = absolute_path + '/folder/my_file.py'

If you want to be operating system agnostic, then you can use:

file_path = os.path.join(absolute_path, folder, my_file.py)

Solution 2:

I have identified the problem. I was running my code on Visual Studio Code debugger. The root directory I have opened was above the level of my file. When I opened the same directory, it worked.