Question

我正在尝试创建一个将消息发送到通道的报告命令。例如* report username＃1234推理放在这里等等等

频道由{作者}报告的{用户}收到{原因}。并向作者发送了一条私人消息，告诉他们报告通过了。

但是，如果所报告的用户具有“踢”权限，则其也会发送到第二个频道。并私下发送给我所有者的消息。与{作者}为{原因}报告的{用户}的结构相同。

但是...我不断收到一些奇怪的错误，例如未定义的bot（即使已定义）。

    async def report(self, ctx, user, *, reason=None):
        logger = bot.get_channel(644966241835941898)
        channel = bot.get_channel(641639710183129113)
        if reason is None:
            await ctx.author.send('Hey I get that you are trying to report someone but I need a reason. Please try again.')
        elif ctx.message.author.guild_permissions.read_messages:
            await ctx.channel.send(f'{user} reported by {author} for {reason}')
        elif user.guild_permissions.kick_members:
            await ctx.logger.send(f'{user}, reported by {author} for {reason}')
            await ctx.channel.send(author, user, reason)
            await ctx.author.send("""Im going to put this in a safe spot for the wizard himself to review.""")```

Answer 1

您需要使用户参数看起来像这样：user : discord.Members，以及要使用户输入之后的所有输入的原因参数。我不确定拥有记录器和频道的意义是什么。

async def report(self, ctx, user : discord.Member, *reason)
    channel = self.bot.get_channel(your channel id) #since it's a cog u need self.bot
    author = ctx.message.author
    rearray = ' '.join(reason[:]) #converts reason argument array to string
    if not rearray: #what to do if there is no reason specified
        await channel.send(f"{author} has reported {user}, reason: Not provided")
        await ctx.message.delete() #I would get rid of the command input
    else:
        await channel.send(f"{author} has reported {user}, reason: {rearray}")
        await ctx.message.delete()

额外

如果您想为有踢烫的人提供一份特殊报告，则if语句将类似于Permissions.kick_members。您可以在api中轻松找到它。

下次再做更多研究，但是如果您需要任何东西，请发表评论：）

使用Discord Python报告命令

1 个答案:

额外