Discord.py 如何在另一个命令中调用另一个命令？

Question

我希望我的机器人在使用已定义的函数 (+play) 键入 +playtest 时播放特定歌曲，但出现错误提示

“Discord.ext.commands.errors.CommandInvokeError：命令引发异常：TypeError：'Command'对象不可调用”

整个代码工作得很好，除了这个命令我想知道 ctx.invoke 是否启用传递参数？或者我只是错过了一些东西

这是我的简短代码

import discord
import wavelink
from discord.ext import commands

import asyncio
from bs4 import BeautifulSoup
import requests
import datetime

queue = []



class Bot(commands.Bot):

    def __init__(self):
        super(Bot, self).__init__(command_prefix=['+'])

        self.add_cog(Music(self))

    async def on_ready(self):
        print(f'Logged in as {self.user.name} | {self.user.id}')


class Music(commands.Cog):

    def __init__(self, bot):
        self.bot = bot

        if not hasattr(bot, 'wavelink'):
            self.bot.wavelink = wavelink.Client(bot=self.bot)

        self.bot.loop.create_task(self.start_nodes())
        self.bot.remove_command("help")

    async def start_nodes(self):
        await self.bot.wait_until_ready()

        await self.bot.wavelink.initiate_node(host='127.0.0.1',
                                              port=80,
                                              rest_uri='http://127.0.0.1:80',
                                              password='testing',
                                              identifier='TEST',
                                              region='us_central')

    @commands.command(name='connect')
    async def connect_(self, ctx, *, channel: discord.VoiceChannel = None):


    @commands.command()
    async def help(self, ctx):

    @commands.command()
    async def play(self, ctx, *, query: str):

    @commands.command(aliases=['sc'])
    async def soundcloud(self, ctx, *, query: str):


    @commands.command()
    async def leave(self, ctx):

    @commands.command(aliases=['queue', 'q'])
    async def check_queue(self, ctx):

    @commands.command(aliases=['clearq', 'clearqueue'])
    async def clear_queue(self, ctx):


    @commands.command(aliases=['removequeue', 'removeq', 'req'])
    async def remove_queue(self, ctx, num: int):

    @commands.command()
    async def skip(self, ctx):


    @commands.command(aliases=['eq'])
    async def equalizer(self, ctx: commands.Context, *, equalizer: str):

    @commands.command()
    async def playtest(self,ctx):
       await ctx.invoke(self.play('hi'))

bot = Bot()
bot.run('sd')

Answer 1

ctx.invoke 确实允许传递参数，但它们需要以与您习惯的方式不同的方式处理 ( function(params))

参数必须在调用中显式显示（例如param = 'value'）并且命令必须是命令对象。这将是您调用命令的方式：

@commands.command()
async def playtest(self, ctx):
   await ctx.invoke(self.bot.get_command('play'), query='hi')