Type 必须满足 VoiceProtocol 抽象基类。 Wavelink/Discord-机器人

Question

我自己无法解决这个问题，我几乎尝试了我能做的一切，我希望有解决办法。

我的 Discord VC-Bot 有以下播放命令，它连接到节点和所有内容。我得到 TypeError: Type must meet VoiceProtocol 抽象基类。 -错误。

目的地是这种类型：

请帮助前辈

@bot.slash_command(guild_ids=[insert ID])
async def play(interaction : nextcord.Interaction, search : str):
    
    query = await wavelink.YouTubeTrack.search(search, return_first=True)

    if not interaction.user.voice or not interaction.user.voice.channel:
        await interaction.response.send_message("You need to be in a voice channel to use this command.")
        return
    

    destination: nextcord.Voicechannel = interaction.user.voice.channel
    
  

    if not interaction.guild.voice_client:

            vc: wavelink.Player = await destination.connect(cls=wavelink.Player)

              #This line breaks my Code and gives me the Type error

            

    else:

            vc: wavelink.Player = interaction.guild.voice_client


    if vc.queue.is_empty and not vc.is_playing():

            await vc.play(query)
            await interaction.response.send_message(f"next disk {vc.current.title}")
    else:
            await vc.queue.put_wait(query)
            await interaction.response.send_message(f"playing track")  

I tried installing pip install -U nextcord[voice]

Answer 1

首先。你能试试吗：

目的地：nextcord.VoiceChannel = interaction.user.voice.channel

然后，如果它仍然不起作用。您应该尝试使用 VS Code 中的

RUN AND DEBUG

并查看给出错误的确切时刻以及变量的值是什么。

如果错误仍然存在并且您想再次寻求帮助，请尽可能发送原始 Traceback。

Answer 2

我希望有完整的追溯？如果我说了一些愚蠢的话，请原谅我，我是一个编码傻瓜，我正在努力学习，但我想我只是把一切都弄错了

所以这里回溯

Traceback (most recent call last):
  File "C:\Users\mydcbot\mydcbot.py", line 60, in play
    vc: wavelink.Player = await destination.connect(cls=wavelink.Player)
  File "C:\Users\.conda\envs\py310\lib\site-packages\nextcord\abc.py", line 1771, in connect
    raise TypeError("Type must meet VoiceProtocol abstract base class.")
TypeError: Type must meet VoiceProtocol abstract base class.