使用函数名作为命令

Question

我这里有这个功能，它使用文件夹名称作为命令。现在仅有的文件夹是“hug”和“headpat”。因此，当我将 .hug 或 .headpat 发送到 Discord 时，它将进入该文件夹，抓取一张随机图像并 post 它。

@client.event
async def on_message(message):
        async def random_image(dir): #picks a random image from local directory
                if message.content.lower() ==f".{dir}":
                   image = random.choice(os.listdir(dir))
            
                   embed = discord.Embed(color = 0xff9e00)
                   chosen_one = discord.File(f"{dir}\{image}", filename = image)

                   embed.set_image(url = f"attachment://{image}")
                   await message.channel.send (embed = embed, file = chosen_one)
                   print(f"{image} sent in response to {(message.author)}.")
                   print(f"----------")

但是，我希望能够执行如下操作，但我不确定从哪里开始。

@bot.command()
async def folder_name(ctx): 
        #code that does the same as the function above

Answer 1

这是一个将您的代码作为命令执行的简单机器人。

import discord
from discord.ext import commands

TOKEN = "your token"
PREFIXES = ['.']
bot = commands.Bot(command_prefix=PREFIXES)


@bot.command(name='image')
async def random_image(ctx, dir):
    image = random.choice(os.listdir(dir))
            
    embed = discord.Embed(color = 0xff9e00)
    chosen_one = discord.File(f"{dir}\{image}", filename = image)

    embed.set_image(url = f"attachment://{image}")
    await ctx.send (embed = embed, file = chosen_one)
    print(f"{image} sent in response to {(message.author)}.")
    print(f"----------")


@bot.event
async def on_ready():
    print(f"Logged in as {bot.user.name} in {len(bot.guilds)} guilds!")


@bot.event
async def on_message(message):
    if message.author.bot: # I always do this to make sure it isn't a bot
       return
    await bot.process_commands(message)


if __name__ == '__main__':
    try:
        bot.run(TOKEN)
    finally:
        print("Logging Out")

Answer 2

默认使用 commands extension, the function's name is the command name. However, the @commands.command decorator have a aliases argument that you can use along with Context.invoked_with:

@bot.command(aliases=['hug', 'headpat'])
async def folder_name(ctx):
    image = random.choice(os.listdir(ctx.invoked_with))
    embed = discord.Embed(color = 0xff9e00)
    chosen_one = discord.File(f"{dir}\{image}", filename=image)

    embed.set_image(url = f"attachment://{image}")
    await ctx.send (embed = embed, file=chosen_one)

    print(f"{image} sent in response to {ctx.author}.")
    print(f"----------")

请注意，如果有人键入 !folder_name，该命令将执行，因此您必须添加如下检查：

if ctx.invoked_with == ctx.command.name:
    #Do stuff

Answer 3

我实际上建议使用 links 而不是 files/folders 因为他们 不可靠 并且经常 会在您的代码中产生问题 和错误，并且它们不能很好地与discord.py 一起工作。这是您所需命令的示例。

import discord
from discord.ext import commands
from discord import Embed
import random
client = commands.Bot(command_prefix=".")

headpat_image_links = ['https://media.tenor.com/images/ad8357e58d35c1d63b570ab7e587f212/tenor.gif', 'https://image.myanimelist.net/ui/xUjg9eFRCjwANWb4t4P8QbyfbVhnxa2QGYxoE7Brq1sHDVak6YxhLP3ZbNkkHm6GX9x8FWB1tWCapNyEqSltXa8wxAdwIERdrHGhYPilHJ8']
secret_headpat = random.choice(headpat_image_links)

@client.command()
async def headpat(ctx):
    e=Embed(title="Headpats!", description="whatever description you want")
    e.set_image(url=secret_headpat)
    await ctx.send(embed=e, content=None)

client.run('TOKEN')

我已经检查过了，效果很好 ;) 这与 hug 命令的概念相同。只需选择一个 random link 嵌入 :)

PS: 确保你没有使用任何 imgur links，imgur 的 API 和 links 已经过时了，不再适用于 discord.py。好吧，它不再适用于嵌入式消息。只是不要使用它。 :>

使用函数名作为命令

use function name as a command

python

function

python-3.x

discord.py

discord.py-rewrite