需要帮助 Discord 机器人队列

我一直在尝试为不和谐机器人和我的>q命令基本上工作为join play queue同时。问题是它只能同时对 2 首歌曲进行排队，所以我需要帮助使其对多首歌曲进行排队

queues = {}
#check queue
def check_queue(ctx, id):
  if queues[id] !=[]:
    voice = ctx.guild.voice_client
    voice.play(queues[id].pop(0))
#command
@client.command(pass_context = True)
async def q(ctx, url):
  if (ctx.author.voice):
    channel = ctx.message.author.voice.channel
    voice = get(client.voice_clients, guild=ctx.guild)
    if voice and voice.is_connected():
      await voice.move_to(channel)
    else:
      voice = await channel.connect()
  else:
    await ctx.send('You are not in a voice channel')      
  YDL_OPTIONS = {'format': 'bestaudio', 'noplaylist': 'True'}
  FFMPEG_OPTIONS = {'before_options': '-reconnect 1 -reconnect_streamed 1 -reconnect_delay_max 5', 'options': '-vn'}
  voice = get(client.voice_clients, guild=ctx.guild)
  with YoutubeDL(YDL_OPTIONS) as ydl:
      info = ydl.extract_info(url, download=False)
  URL = info['url']
  source = (FFmpegPCMAudio(URL, **FFMPEG_OPTIONS))
  if voice.is_playing():
    guild_id = ctx.message.guild.id
    if guild_id in queues:
      queues[guild_id].append(source)
    else:
      queues[guild_id]=[source]
    await ctx.send('Deemo: Song added to queue')
  else:
      ctx.voice_client.stop()
      voice.play(FFmpegPCMAudio(URL, **FFMPEG_OPTIONS), after=lambda x=0: check_queue(ctx, ctx.message.guild.id))
      voice.is_playing()
      await ctx.send('Playing: '+ info.get('title'))

在我做了一些研究和搜索之后，我找到了一种简单的方法来使队列循环 2+ 首歌曲，添加after=lambda x=0: check_queue(ctx, ctx.message.guild.id) inside check_queue稍微改变一下，它会看起来像这样

def check_queue(ctx, id):
  if queues[id] !=[]:
    voice = ctx.guild.voice_client
    source = queues[id].pop(0)
    voice.play(source, after=lambda x=0: check_queue(ctx, ctx.message.guild.id))