Python Flask Frozen 没有属性 'url_default_functions'

Question

I have written my 1st Python API project with Flask and now am trying to deploy it to Netlify.

Searched online and found I need to use Flask-Frozen to generate a static website. Not sure I'm doing it correctly, because my project is a API project not a website project, so may be I should not use Flask-Frozen(FF)?

But if I could still use FF to generate static website for my API project, here is my project:

--

• app.py
• mazesolver
  • mazeapi.py

Here is the app.py

from flask_frozen import Freezer
from mazesolver import mazeapi

# Call the application factory function to construct a Flask application
# instance using the development configuration
# app = mazeapi()

# Create an instance of Freezer for generating the static files from
# the Flask application routes ('/', '/breakfast', etc.)
freezer = Freezer(mazeapi)


if __name__ == '__mazeapi__':
    # Run the development server that generates the static files
    # using Frozen-Flask
    freezer.run(debug=True)

mazeapi.py

import io
from mazesolver.solver import MazeSolver
from markupsafe import escape
from flask import Flask, flash, request, redirect, send_file
from werkzeug.utils import secure_filename
    
ALLOWED_EXTENSIONS = { 'png', 'jpg', 'jpeg' }

app = Flask(__name__)
app.config['MAX_CONTENT_LENGTH'] = 5 * 1024 * 1024

@app.route('/maze/<mazename>')
def maze(mazename):
    return 'maze 4 %s' % escape(mazename)

Whenever I run: python app.py

I got this error:

Traceback (most recent call last):
  File "app.py", line 10, in <module>
    freezer = Freezer(mazeapi)
  File "/home/winstonfan/anaconda3/envs/maze/lib/python3.7/site-packages/flask_frozen/__init__.py", line 98, in __init__
    self.init_app(app)
  File "/home/winstonfan/anaconda3/envs/maze/lib/python3.7/site-packages/flask_frozen/__init__.py", line 108, in init_app
    self.url_for_logger = UrlForLogger(app)
  File "/home/winstonfan/anaconda3/envs/maze/lib/python3.7/site-packages/flask_frozen/__init__.py", line 588, in __init__
    self.app.url_default_functions.setdefault(None, []).insert(0, logger)
AttributeError: module 'mazesolver.mazeapi' has no attribute 'url_default_functions'

Answer 1

I had the same issue and added

url_default_functions={}

to the app, right before

if __name__ == "__main__":

    app.run()

I guess you could put it right after

app = Flask(__name__)

and the error went way... but there are many more:) It seems that frozen-flask isn't working with the lastest flask, but I couldn't get any other versions working with it.

Did you manage to get it working?

Answer 2

You have a simple namespace bug.

In the code you've posted, you are freezing your module mazeapi , not the actual Flask application, mazeapi.app .

Try changing your Freezer call in line 10 of app.py to:

freezer = Freezer(mazeapi.app)