Flask - How to use CSS when i'm using Response(stream_template) with generator

Question

I'm trying to learn Flask and making a small app. So at first, I tested without a css file by using: (delay() gets the result from generator)

return Response(stream_template('login.html', data=delay()))

It works fine for me then now i want to implement new css, let's called it style.css and i put it in the static folder. In the html file I have:

<link rel="stylesheet" href="{{ url_for('static', filename='style.css') }}"/>

The code will probably won't work as there is a problem with context and response, but it will work fine as a static return:

return render_template('login.html')

My Question is it it anyway that I can both have the generator to work (the delay() function) with the CSS in the static folder? I just spent couple hours on this problem but couldn't find an answer yet.

My stream_template:

def stream_template(template_name, **context):
    app.update_template_context(context)
    t = app.jinja_env.get_template(template_name)
    rv = t.stream(context)
    # uncomment if you don't need immediate reaction
    ##rv.enable_buffering(5)
    return rv

Thanks a lot

Answer 1

Quoting from the documentation:

Note that when you stream data, the request context is already gone the moment the function executes. Flask 0.9 provides you with a helper that can keep the request context around during the execution of the generator: ...

In your case, your code probably should be:

return Response(stream_with_context(stream_template('login.html', data=delay())))

or

return Response(stream_template('login.html', data=stream_with_context(delay())))