Flask.url_for() error: Attempted to generate a URL without the application context being pushed

Question

I have a trivial app where I\'m trying to redirect the favicon per:

http://flask.pocoo.org/docs/0.10/patterns/favicon/

app = flask.Flask(__name__)
ap         


        

Answer 1

Late answer, I've just run into the same problem. I don't see a big downside in handling the redirect like this instead:

@app.route('/favicon.ico')
def favicon():
    return redirect(url_for('static', filename='favicon.ico'))

This prevents url_for from being called before the application is ready.

To give a counterpoint to using a link in the HTML only, it's a good practice for every site to have a favicon.ico and robots.txt at the root level - even if they're empty. It avoids problems like this and other unnecessary errors that adds noise to logs.