Question

我正在用不和的手法制造一个新的不和谐的机器人,我希望与使用我的机器人的人交往。例如,我将打上“/mention @myfirend”的字句,然后由我的机器人回答“Lexspe提到@myLG”。但是,如果没有提及的话,机器人将回答“你应当提及”的问题。

我没有这样的想法,因此我要求翻案,并说,如果你想要用刀子来做,你应该使用不同的图书馆,但我对其他图书馆没有想法,因此我再次问,复印机开始提供 st的回答,其中有许多错误,我想在这里问你。

此外,我还搜索了许多论坛,但所有论坛都不是关于闪电的论坛,因此我仍然不知道如何核对信息。

我的守则如下:

import discord
from discord import app_commands
from discord.ext import commands

bot = commands.Bot(command_prefix="!", intents=discord.Intents.default(), activity=discord.Game(name= with arda ))

@bot.event
async def on_ready():
    print("started")
    try:
        synced = await bot.tree.sync()
        print(f"Synced {len(synced)} commands(s)")
    except Exception as e:
        print(e)

@bot.tree.command(name = "gizli")
async def sa(interaction: discord.Interaction):
    await interaction.response.send_message(f"as {interaction.user.mention}, slash komudu calisti ezzzzz", ephemeral=True)

@bot.tree.command(name = "ping")
@app_commands.describe(kimi = "say something")
async def say(interaction: discord.Interaction, kimi: str):
    await interaction.response.send_message(f"{interaction.user.mention} mentioned {kimi}")

@bot.tree.command(name = "say")
@app_commands.describe(message = "say something")
async def say(interaction: discord.Interaction, message: str):
    await interaction.response.send_message(f"{interaction.user.mention} said {message}")

bot.run("token")

我做了一些这样的事情,但正如我所说的那样,当有人使用“/选择”时,我怎么能够核对信息?

Answer 1

You could just typehint kimi as a discord.Member object and if the argument is not provided the bot answers with "You should mention someone", otherwise the bot will mention the provided member.

@bot.tree.command(name = "ping")
@app_commands.describe(kimi = "say something")
async def say(interaction: discord.Interaction, kimi: discord.Member=None):
    if kimi is None:
       await interaction.response.send_message("You should mention someone!")
    await interaction.response.send_message(f"{interaction.user.mention} mentioned {kimi.mention}")

Answer 2

我无法对此进行测试,因此我可以肯定会这样做,但我想到的是:

@bot.tree.command(name = "ping")
@app_commands.describe(kimi = "say something")
async def say(interaction: discord.Interaction):
    if interaction.message.mentions: # checks that the list of mentions is not empty
        toSend = f" mentionned {interaction.message.mentions[0]}"
    else:
        toSend = " did not manage to mention anyone"
    await interaction.response.send_message(f"{interaction.user.mention} {toSend}")

为了检查引发激烈指挥的信息的内容是否含有提及的内容,你需要检索互动的归属信息。然后,你可以使用该电文的属性,看看该电文是否有任何有效的提及。

Interaction Object Documentation

Answer 3

不要忘记该成员的物体,如@boez的话,你检查了即将到来的参数,看看它是否有有效的用户支持方。例如:

<@user-id>

以上是适当的@(有些人)应该看。如果我们把用户带在单版上,我们就可以这样做。如果你想把某些用户排除在外,这样做也是有益的。

当我们用以下法典印刷上述文字时,我们就可以看到这一点:

@tree.command(name="say", guilds=[discord.Object(id=no)])
@app_commands.describe(user="Input valid user @")
async def say(interaction: discord.Interaction, user: str):
    print(user)  # print the user
    await interaction.response.send_message(f"{interaction.user.mention} mentioned {user}")

这一回报:

<@497930397985013781>

如果我们编辑该守则以检查该身份证是否有效:

@tree.command(name="say", guilds=[discord.Object(id=no)])
@app_commands.describe(user="Input valid user @")
async def say(interaction: discord.Interaction, user: str):
    user_id = re.sub("[^0-9]", "", user)  # remove all non characters
    if user_id.isdigit():  # check if it s a digit
        if interaction.guild.get_member(int(user_id)) is not None:  # String splice the mention to get the ID and pass it to get_member
            return await interaction.response.send_message(f"{interaction.user.mention} mentioned {user}")

    await interaction.response.send_message(f"Not a mention")

以上使用<条码>re,但仅凭<条码>进口<>。如果你不希望利用,你就只会使扼杀。

产出:

我对我说真话时,我发出最崇高的信息。底线是我发出非@讯息。

友情链接