需要帮助 Discord 机器人队列 [英] Need help Discord bot queue

问题描述

我一直在尝试为不和谐机器人创建队列，而我的 >q 命令基本上可以作为 joinplay queue 同时进行.问题是它只能同时排队 2 首歌曲，所以我需要帮助让它排队多首歌曲

i've been tryin to make a queue for discord bot and my >q command basicaly work as join play queue at the same time. The problem is it only queues 2 songs at the same time so I need help making it queue multi songs

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'))

推荐答案

经过一番研究和搜索，我想出了一个简单的方法来让队列循环 2+ 首歌曲，添加 after=lambda x=0:check_queue(ctx, ctx.message.guild.id) 在 check_queue 里面稍微改一下，就变成这样了

after I do some study and search I figure out an easy way to make the queue loop 2+ songs, add after=lambda x=0: check_queue(ctx, ctx.message.guild.id) inside check_queue and change a little and it will look like this

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))

这篇关于需要帮助 Discord 机器人队列的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！