My directory structure like below:
./outputsetting.json
./myapp/app.py
I load outputsetting.json file in app.py like below:
with open("..\outputpath.json","r") as f:
j=json.load(f)
And run in myapp directory it’s ok:
python app.py
But if I run app.py in the parent directory:
python .\myapp\app.py
It raise error
FileNotFoundError: [Errno 2] No such file or directory: '..\\outputpath.json'
How can I load file from disk by the relative directory to the app.py? So I can run it from any place and needn’t modify code every time.
Thanks!
>Solution :
When you start the script from the parent directory, the working directory for the script changes. If you want to access a file that has a specific location relative to the script itself, you can do this:
from pathlib import Path
location = Path(__file__).parent
with open(location / "../outputsetting.json","r") as f:
j=json.load(f)
Path(__file__) gets the location of the script it is executed in. The .parent thus gives you the folder the script is in, still as a Path object. And you can then navigate to the parent directory from that with location / "../etc"
Or of course in one go:
with open(Path(__file__).parent / "../outputsetting.json","r") as f:
j=json.load(f)
(Note: your code says outputpath.json, but I assume that’s the same outputsetting.json)
Another solution would be to just change the working directory to be the folder the script is in – but that of course only works if all your scripts and modules are OK with that change:
from pathlib import Path
from os import chdir
chdir(Path(__file__).parent)
with open("../outputsetting.json","r") as f:
j=json.load(f)
I’d prefer constructing the absolute path as in the previous example though.