Flask: How To Read A File In Application Root?

My Flask application structure looks like application_top/ application/ static/ english_words.txt templat

Solution 1:

I think the issue is you put / in the path. Remove / because static is at the same level as views.py.

I suggest making a settings.py the same level as views.py Or many Flask users prefer to use __init__.py but I don't.

application_top/
    application/
          static/
              english_words.txt
          templates/
              main.html
          urls.py
          views.py
          settings.py
    runserver.py

If this is how you would set up, try this:

#settings.py
import os
# __file__ refers to the file settings.py 
APP_ROOT = os.path.dirname(os.path.abspath(__file__))   # refers to application_top
APP_STATIC = os.path.join(APP_ROOT, 'static')

Now in your views, you can simply do:

import os
from settings import APP_STATIC
withopen(os.path.join(APP_STATIC, 'english_words.txt')) as f:
    f.read()

Adjust the path and level based on your requirement.

Solution 2:

Here's a simple alternative to CppLearners answer:

from flask import current_app

with current_app.open_resource('static/english_words.txt') asf:
    f.read()

See the documentation here: Flask.open_resource

Solution 3:

The flask app also has a property named root_path to resolve the root directory as well as an instance_path property for the particular app directory without requiring the os module, though I like @jpihl's answer.

withopen(f'{app.root_path}/static/english_words.txt', 'r') as f:
    f.read()