我有这个函数
,可以播放Spotify
:
#a function to play Spotify
def play(id_):
print 'playing', id_
os.system("osascript -e 'tell application \"Spotify\" to play track \"%s\"'" % (id_,))
接下来的循环
会遍历所有播放列表
歌曲,获取所有可播放的id
(foreign_id
),将它们传递给 play(id_)
,
并将每首歌曲duration
传递给time.sleep()
以停止循环,直到每首歌曲结束,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,只播放一首歌,递归就消失了。
如何更正代码,以便让函数按顺序播放所有轨道,仅运行代码一次?
最佳答案
看起来 play(track_id)
应该位于 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
关于python - 如何设置计时器来控制循环内的函数调用?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/36991241/