[FIXED] What is the correct way to serve/access media files in Flask

Issue

I am struggling with the following problem: what is the correct way to serve media files in Flask? I have build a model Posts in a blog system, I'm using flask-file-upload

# models/blog.py
from app import db, file_upload

@file_upload.Model
class Posts(db.Model):
    """
    Defines the attributes of posts table in the database
    """
    ...
    image = file_upload.Column()
    ...

in my settings.py file i have

import os
from pathlib import Path

PROJECT_ROOT = Path(__file__).parent.parent
UPLOAD_FOLDER = os.path.join(PROJECT_ROOT, 'media')

so, when I add a Post object it saves the file name in the database and the file in media folder with path media/posts/<post_id>/filename

So far it works perfectly. But I need serve these images, so I created the following route:

# routes.py

from app import app

@app.get('/media/<path:path>')
def send_media(path):
    """
    :param path: a path like "posts/<int:post_id>/<filename>"
    :return:
    """
    path_list = path.split('/')
    path_ = '/'.join(path_list[:-1]) + '/'
    file_name = path_list[-1]
    return send_from_directory(
        directory=app.config['UPLOAD_FOLDER'], path=path_, filename=file_name, as_attachment=True
    )

but it is not working, when I browse http://127.0.0.1:5000/media/posts/15/python.jpg

my Flask app responds with 404 - Not Found, even the python.jpg file being there.

How can I correctly access the media files in Flask? I need this because I have to create an API that sends the file_name and the frontend will render the image, something like

<img src="http://localhost:8000/media/posts/<post_id>/file_name">

Solution

Solution: seeing the send_from_directory() docstring I got it! We don't need path_list, path_ and file_name variables. The correct way to serve media files with Flask is:

@app.get('/media/<path:path>')
def send_media(path):
    """
    :param path: a path like "posts/<int:post_id>/<filename>"
    """

    return send_from_directory(
        directory=app.config['UPLOAD_FOLDER'], path=path
    )

Answered By - Jailton Silva