CODEWITHSUNDEEP
pythonworking-directoryproject-folder
rams
rams

Reputation: 128

Python is throwing error for a file path inside the Project folder

Python is throwing some inconsistent error for a file reference inside the Project folder based on 'Working Directory' in the script configuration My Project Structure is

My Project Structure as follows

config_utils.py

f = ''

def config_init():

    global f

    txt_dir = '../files/sample.txt'

    f = open(txt_dir, "r")

    f = f.read()

mycode.py

import config_ru.config_utils as cu

cu.config_init()

print(cu.f)

On executing mycode.py, it throws the below error w.r.t "sample.txt" in "files" package

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but if I change the Working directory of "my_code.py" in the script configuration from "level2" to "level1", mycode.py gets executed successfully

enter image description here

enter image description here

This is very weird because in both the cases the location of "sample.txt" remains unchanged and both the error and being forced to change the Working Directory seems to be unacceptable. Please clarify

Upvotes: -1

Views: 848

Answers (2)

quamrana
quamrana

Reputation: 39404

The work-around is to get the path of the module you are in and apply the relative path of the resource file to that:

from pathlib import Path

f = ''

def config_init():
    global f

    p = Path(__file__).parent.absolute()
    txt_dir = (p / '../files/sample.txt').resolve()

    f = open(txt_dir, "r")
    f = f.read()

Upvotes: 2

AndrzejO
AndrzejO

Reputation: 1582

Looks like normal behavior. In the line

txt_dir = '../files/sample.txt'

the .. means 'go one directory up'. So, when you are in level2, it will go up one level (to level1) and look for files/sample.txt, which does not exist. However, when you are in level1, then the .. will bring you to the pythonProject dir, where it can find files/sample.txt

Upvotes: 1

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