Python directory file go wrong

Question

I am having a silly problem with reading a file in python.

I have a folder 'a' that contains my test.py and file test.json for reading.

My test.py looks like this:

config_path = 'test.json'
with open(config_path) as config_buffer:
    config = json.loads(config_buffer.read())

When I go outside the directory structure of folder 'a' and I run the command:

python a\test.py

And the console returns this error:

FileNotFoundError: No such file or directory: 'test.json'

I try using the absolute file path using pathlib:

config_path = Path('end2end.json').absolute()
with open(config_path) as config_buffer:
    config = json.loads(config_buffer.read())

But it still returns this error to me:

FileNotFoundError: No such file or directory: 'D:\\test.json'

Can anyone help me to get exactly the right directory file?

try your first code, but run it from inside the directory, not from the outside — Gal, May 18 '20 at 09:43
It looks like this is an issue with the "current directory". You might find this post helpful: https://stackoverflow.com/questions/5137497/find-current-directory-and-files-directory — Tobias, May 18 '20 at 09:46

score 1 · Accepted Answer · answered May 18 '20 at 09:59

If you want to call your python script from any folder and let it access its local files without specifying paths, you can change the directory in the begining:

import os
os.chdir(os.path.dirname(os.path.realpath(__file__)))

config_path = 'test.json'
...

score 0 · Answer 2 · answered May 18 '20 at 09:59

The Problem is that you should place the test.json file in the current working directory.

i.e)For Example, The python file is located in C:\users\Desktop\Code\test.py then you should copy the file into C:\users\Desktop\Code\

(OR)

You should give the path of the file in with open($PATH TO JSON FILE) as config_buffer

Python directory file go wrong

2 Answers2