Can't open file in the same directory

Question

I was following a python tutorial about files and I couldn't open a text file while in the same directory as the python script. Any reason to this?

f = open("test.txt", "r")

print(f.name)

f.close()

Error message:

Traceback (most recent call last):
  File "c:\Users\07gas\OneDrive\Documents\pyFileTest\ManipulatingFiles.py", line 1, in <module>
    f = open("test.txt", "r")
FileNotFoundError: [Errno 2] No such file or directory: 'test.txt'

Here's a screenshot of proof of it being in the same directory:

It's telling you that `test.txt` isn't in the current working directory. We don't know what that is. How are you running the code, and what are the environment options? — Peter Wood, Nov 30 '21 at 19:52
Replace it with this. First `import os` then `f = open(os.path.abspth("test.txt"), "r")` and see what error you get. — Peter Wood, Nov 30 '21 at 19:53
Im running the code through Visual Studio Code I'm not sure what you mean by environment — Geomatrix Gamer, Nov 30 '21 at 19:53
I this error : ``` File "c:\Users\07gas\OneDrive\Documents\pyFileTest\ManipulatingFiles.py", line 3, in f = open(os.path.abspth("test.txt"), "r") AttributeError: module 'ntpath' has no attribute 'abspth'. Did you mean: 'abspath'?``` — Geomatrix Gamer, Nov 30 '21 at 19:57
well use `abspath` then, clearly (based on the error) it was a typo on their part; code block formatting is available only in posts; could you also try `print(os.getcwd())` and see what you get — Matiiss, Nov 30 '21 at 19:58
I got this error after fixing typo : `File "c:\Users\07gas\OneDrive\Documents\pyFileTest\ManipulatingFiles.py", line 3, in f = open(os.path.abspath("test.txt"), "r") FileNotFoundError: [Errno 2] No such file or directory: 'C:\\Users\\07gas\\test.txt'` — Geomatrix Gamer, Nov 30 '21 at 20:00
@GeomatrixGamer well, that error clearly shows that python is trying to find the file in `07gas` directory, which means that the interpreter runs there — Matiiss, Nov 30 '21 at 20:01
that would be an immediate solution but it depends on how the python script is run exactly because you can definitely make it run from the same directory as the script, at the very least by using `open(os.path.join(os.path.dirname(__file__), 'test.txt'), 'r')` — Matiiss, Nov 30 '21 at 20:03
See this question and answer: https://stackoverflow.com/questions/38623138/vscode-how-to-set-working-directory-for-debug — Peter Wood, Nov 30 '21 at 20:52
Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. — Community, Dec 05 '21 at 17:42

martineau · Answer 1 · 2021-11-30T20:44:42.367

The problem is "test.txt" is a relative file path and will be interpreted relative to whatever the current working directory (CWD) happens to be when the script is run. One simple solution is to use the predefined __file__ module attribute which is the pathname of the currently running script to obtain the (aka "parent") directory the script file is in and use that to obtain an absolute filepath the data file in the same folder.

You should also use the with statement to ensure the file gets closed automatically.

The code below shows how to do both of these things:

from pathlib import Path

filepath = Path(__file__).parent / "test.txt"

with open(filepath, "r") as f:
    print(f.name)

Can't open file in the same directory

1 Answers1