【发布时间】:2021-08-06 06:11:42
【问题描述】:
在使用这行代码时返回成员 ID 时遇到问题,当我不调用任何东西时,我只是让它运行为:
table = ("\n".join(f'{idx + 1}. {(entry[0])} (XP: {entry[1]} | Level: {entry[2]})'
for idx, entry in enumerate(entries)))
我将用户 ID 存储在我的数据库中。
但是一旦我尝试将它们转换为不和谐的成员用户名,我就会得到非类型对象
table = ("\n".join(f'{idx + 1}. {self.ctx.guild.get_member(entry[0])} (XP: {entry[1]} | Level: {entry[2]})'
for idx, entry in enumerate(entries)))
需要更多代码来回答这个问题,所以在这里
class HelpMenu(ListPageSource):
def __init__(self, ctx, data):
self.ctx = ctx
super().__init__(data, per_page = 10)
async def write_page(self, menu, fields=[]):
offset = (menu.current_page * self.per_page) + 1
len_data = len(self.entries)
embed = Embed(title="Server XP Leaderboard",
colour=self.ctx.author.colour)
embed.set_thumbnail(url = self.ctx.guild.icon_url)
embed.set_footer(text = f"{offset:,} - {min(len_data, offset+self.per_page-1):,} of {len_data:,} members.")
for name, value in fields:
embed.add_field(name=name, value=value, inline=False)
return embed
async def format_page(self, menu, entries):
fields = []
table = ("\n".join(f'{idx + 1}. {self.ctx.guild.get_member(entry[0])} (XP: {entry[1]} | Level: {entry[2]})'
for idx, entry in enumerate(entries)))
fields.append(("Ranks", table))
return await self.write_page(menu, fields)
class Exp(Cog):
def __init__(self, bot):
self.bot = bot
@command(aliases = ['lvl', 'level'])
async def rank(self, ctx):
db = sqlite3.connect('xpdata.db')
cursor = db.cursor()
# user_id = self.author.id
guild_id = ctx.guild.id
#
# cursor.execute(f'SELECT * FROM xpdata WHERE user_id = {user_id} AND guild_id = {guild_id}')
# for row in cursor.fetchall():
# level_display = row[3]
cursor.execute(f'SELECT user_id, xp, level FROM xpdata WHERE guild_id = {guild_id} ORDER BY xp DESC')
xp_ranking = cursor.fetchall()
#menu
ranking_menu = MenuPages(source=HelpMenu(ctx, xp_ranking))
await ranking_menu.start(ctx)
#await ctx.channel.send('{} is currently level {} and rank {}'.format(ctx.author.mention, level_display))
bot.add_cog(Exp(bot))
【问题讨论】:
-
你有所有的意图吗?
-
我对你的意思感到困惑
-
你有不和谐的特权意图吗?机器人需要识别服务器中的成员。 discordpy.readthedocs.io/en/stable/intents.html
-
我刚刚启用了它,但我仍然收到同样的错误
-
另外,我认为你应该使用
ctx.guild.fetch_member('id here')而不是ctx.guild.get_member(entry[0])。对于您的代码,是否定义了self.ctx?即使这个命令在一个类中,我认为你应该有像async def command_name(self, ctx)这样的参数