我有一个处理后台进程的Flask视图,如下所示:
@app.route('/playlist')
def playlist(tracks):
executor.submit(build_playlist,tracks=tracks)
return 'Building playlist...'
并且此视图函数调用以下方法:
def build_playlist(tracks):
(...process tracks...)
这是在应用程序中调用该进程的地方:
@app.route('/handle_playlists', methods=['POST'])
def handle_playlist():
jukebox = ['track1', 'track2', 'track3']
tracks = random.choice(jukebox)
playlist(tracks) <------------------------
如果我致电playlist()
,我会收到错误消息:
`'unicode' object is not callable`
但如果我直接致电build_playlist(tracks)
,它就会有用,所以我必须在executor.submit()
犯一个愚蠢的错误。
我在这里错过了什么?