Flask: How to Read a File in Application Root?
My Flask Application Structure Looks Like Application_Top/ Application/ Static/ English_Words. Txt Templates/ Main. Html Urls. Py Views. Py Runserver. Py When...
My Flask application structure looks like
application_top/
application/
static/
english_words.txt
templates/
main.html
urls.py
views.py
runserver.py
When I run the runserver.py, it starts the server at localhost:5000.
In my views.py, I try to open the file english.txt as
f = open('/static/english.txt')
It gives error IOError: No such file or directory
How can I access this file?
3 Answers
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
with open(os.path.join(APP_STATIC, 'english_words.txt')) as f:
f.read()
Adjust the path and level based on your requirement.
Here's a simple alternative to CppLearners answer:
from flask import current_app
with current_app.open_resource('static/english_words.txt') as f:
f.read()
See the documentation here: Flask.open_resource
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.
with open(f'{app.root_path}/static/english_words.txt', 'r') as f:
f.read()