如何设置计时器来控制循环内的函数调用？

Posted 2023-04-15

【中文标题】如何设置计时器来控制循环内的函数调用？

【英文标题】：How can I set a Timer to control a function call inside a loop?

【发布时间】：2016-08-27 17:42:17

【问题描述】：

我有这个function，它播放Spotify：

#a function to play Spotify
def play(id_):
    print 'playing', id_
    os.system("osascript -e 'tell application \"Spotify\" to play track \"%s\"'" % (id_,))

以及下面的loop，遍历所有playlist的歌曲，获取所有可播放的id的（foreign_id），传递给play(id_)，

然后将每首歌曲 duration 传递给 time.sleep() 以停止循环，直到每首歌曲完成，然后再次重复循环：

for i, song in enumerate(song_playlist):
            #we need to track each song id
            song_id = song_playlist[i]['id']
            #in order to get song 'duration', access 'song/profile response' and pass the id as an argument
            response_profile = en.get('song/profile', id=song_id, bucket="audio_summary")
            song_profile = response_profile['songs']
            dur = song_profile[0]['audio_summary']['duration']
            #convert to miliseconds     
            dur *= 1000
            print int(round(dur))                           
            #now we access each song 'foreign_id' 
            for track in song:
                track = song['tracks'][i]
                track_id = track['foreign_id'].replace('-WW', '')           
            print '0 2 1'.format(i, song['artist_name'], song['title'])
            #call the function for each track
            play(track_id) #CALL FUNCTION HERE
            time.sleep(int(round(dur))) # SET INTERVAL CALL TO EACH SONG DURATION

但是，只播放了一首歌曲，递归消失了。

如何更正代码，以便我可以让按顺序播放所有曲目的功能只运行一次代码？

【问题讨论】：

不能通过song_id = song['id']获得song_id

【参考方案1】：

看起来play(track_id) 应该在for track in song 循环内。您需要将其缩进 1 级。

for i, song in enumerate(song_playlist):
    # Code as before ...
    for track in song:
        track = song['tracks'][i]
        track_id = track['foreign_id'].replace('-WW', '')           
        print '0 2 1'.format(i, song['artist_name'], song['title'])
        play(track_id) #CALL FUNCTION HERE
    time.sleep(int(round(dur))) # SET INTERVAL CALL TO EACH SONG DURATION