Python sqlite查询中缺少len()
我一直在尝试根据经验对用户进行排名,尽管在调用命令rank时,我似乎收到了缺少len()的类型错误 错误:命令引发异常:TypeError:类型为“sqlite3.Cursor”的对象没有len() 我正在调用一个帮助类(虽然这也是我第一次使用类),在它里面,它调用len,这给了我一个错误 谢谢你的帮助Python sqlite查询中缺少len(),python,sqlite,discord.py,typeerror,Python,Sqlite,Discord.py,Typeerror,我一直在尝试根据经验对用户进行排名,尽管在调用命令rank时,我似乎收到了缺少len()的类型错误 错误:命令引发异常:TypeError:类型为“sqlite3.Cursor”的对象没有len() 我正在调用一个帮助类(虽然这也是我第一次使用类),在它里面,它调用len,这给了我一个错误 谢谢你的帮助 class HelpMenu(ListPageSource): def __init__(self, ctx, data): self.ctx = ctx
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 = []
for idx, entry in enumerate(entries):
fields.append(("Ranks", "\n".join(f'{idx}. {entry[0]} (XP: {entry[1]} | Level: {entry[2]})')))
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]
xp_ranking = cursor.execute(f'SELECT user_id, xp, level FROM xpdata WHERE guild_id = {guild_id} ORDER BY xp DESC')
#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))
您没有获取查询的结果 更正此行:
xp\u ranking=cursor.execute(f'SELECT user\u id,xp,level FROM xpdata,其中guild\u id={guild\u id}按xp DESC'排序)
到
cursor.execute(f'SELECT user_id,xp,levelfrom xpdata,其中guild_id={guild_id}按xp DESC'排序)
xp_ranking=cursor.fetchall()
@Barmar哦,是的,你说得对。我想在一行中变得聪明。。。坏主意——。-(我编辑了我的帖子)不幸的是,流畅的界面不是Python的标准。