I am writing a flask app that I will deploy it on Heroku to test it. There is a page that will generate daily photos and information based on a JSON file. The view is valid for 24 hours beginning from 00:00 to 23:59 of everyday.I am thinking about using a cache, otherwise the app will consume unnecessary resources.import ..TODAY = date.today() app = Flask(__name__) app.jinja_env.auto_reload = False cache = Cache(config={'CACHE_TYPE': 'simple', 'CACHE_ARGS' : TODAY})#This function reads a json file, filter the daily data to process and render a html template with photos and text .# 3600 *  24 = 86400@cache.cached(timeout=86400)@app.route('/inf')def inf():    input_file = open(os.path.join(__location__, 'json/output.json'));        j = json.loads(input_file.read().decode(utf-8-sig))    etc ..    return render_template('inf.html',                         title = title,                        config={'environment_args':{'autoescape':False}},                        zipped =zipped) if __name__ == '__main__':    if TODAY != cache.config['CACHE_ARGS'] :        with app.app_context():            cache.clear()                    port = int(os.environ.get('PORT', 33507))    app.run(debug = True, host='0.0.0.0', port=port)I am not sure if this is the best solution to cache the view for a day.  That's why I am asking here for more suggestions.
Caching a function view for a day
python;cache;flask