不和谐中缺少参数消息

Question

所以我用不同的帮助菜单制作了一个帮助命令。就像 mee6 一样。但我想添加一条消息，因为没有给出任何参数。怎么做？这是我现在拥有的：

@bot.command(name='help')
async def help(ctx, *, content):
    if content == ('Moderation'):
        await ctx.send(moderationmenu)
    if content == ('fun'):
        await ctx.send(funmenu)
    if content == None:
        await ctx.send('please provide an argument (Moderation / fun)')

Answer 1

您可以像这样提供默认 arg 值：

@bot.command(name="help")
async def help(ctx, *, content = None):
    if not content: # more pythonic way of checking a variable is None or not
        await ctx.send("Please provide an argument (moderation / fun)")
    elif content.lower() == "fun": # brackets not necessary
        await ctx.send(funmenu)
    elif content.lower() == "moderation": # makes it case insensitive
        await ctx.send(moderationmenu)
    else:
        await ctx.send("Sorry, I didn't recognise that category. Please choose (moderation / fun)")

快速编辑 - 如果您将菜单分类为单独的变量，则可以将它们映射到字典中：

@bot.command(name="help")
async def help(ctx, *, content = None):
    menus = {"fun": funmenu, "moderation": moderationmenu}
    if not content:
        await ctx.send("Please provide an argument (moderation / fun)")
    else:
        try:
            await ctx.send(menus[content.lower()])
        except KeyError:
            await ctx.send("Sorry, I didn't recognise that category. Please choose (moderation / fun)")

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参考资料：

• Default args