我正在构建一个Flask需要后台进程导致上传到SQLAlchemy数据库的应用程序。相关片段:from flask_sqlalchemy import SQLAlchemyimport concurrent.futuresimport queuefrom models import Upload_Tracksapp = Flask(__name__)db.init_app(app)app.config.update(SQLALCHEMY_DATABASE_URI= "sqlite:///%s" % os.path.join(app.root_path, 'path/to/player.sqlite3'))q = queue.Queue()在database.py:from flask_sqlalchemy import SQLAlchemydb = SQLAlchemy()在models.py:def Upload_Tracks(item): uri = None title = unidecode(item['title']) artist = unidecode(item['artist']) preview = item['preview'] energy = item['energy'] popularity = item['popularity'] tempo = item['tempo'] brightness = item['brightness'] key = item['key'] image = item['artist_image'] duration = item['duration'] loudness = item['loudness'] valence = item['valence'] genre = item['genre'] track = Track(title=title, artist=artist, uri=uri, track_id=None) db.session.add(track) track.preview = preview track.energy = energy track.popularity = popularity track.tempo = tempo track.genre = genre track.brightness = brightness track.key = key track.image = image track.duration = duration track.loudness = loudness track.valence = valence db.session.commit()数据库工作正常,没有线程。但是,当我运行它时,会捕获以下异常:co/mp3-preview/c0d57fed887ea2dbd7f69c7209adab71671b9e6e?cid=d3b2f7a12362468daa393cf457185973 '} 生成了一个异常:未找到应用程序。在视图函数内工作或推送应用程序上下文。请参阅http://flask-sqlalchemy.pocoo.org/contexts/。但据我所知,这个过程是在@app.route. 这怎么断章取义了?我该如何解决?
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