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

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隐瞒了意图╮
隐瞒了意图╮ 2020-12-16 10:53

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
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3条回答
  • 春和景丽
    2020-12-16 11:01

    Don't put this in the app but in the html file

        <html lang="en">
    <head>
        <title>{{ title }}</title>
         <meta charset="utf-8">
         <meta name="viewport" content="width=device-width/2, initial-scale=1">
         <link rel="stylesheet" href="{{ url_for('static', filename='css/style.css') }}">
    </head>
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  • 予麋鹿
    2020-12-16 11:04

    According to the doc:

    Setting a SERVER_NAME also by default enables URL generation without a request context but with an application context.

    since you're using app_context, you may set the SERVER_NAME Configuration Value.

    By the way, as the doc:Adding a favicon says:

    <link rel="shortcut icon" href="{{ url_for('static', filename='favicon.ico') }}">

    the above line should be enough for most browsers, we don't have to do any other things.

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  • 孤城傲影
    2020-12-16 11:15

    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.

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