list of users in embed
@bot.command(name="list")
async def __sk(ctx, frac: int = None, dos: int = None):
if frac is None:
for value in sql.execute(f"SELECT idds, dostup FROM users"):
name, dostup = value
member = bot.get_user(name)
print(member)
await ctx.send(embed=discord.Embed(description=f" Пользователь {member.mention} | Доступ - {dostup}"))
else:
if dos is None:
for value in sql.execute(f"SELECT idds, dostup FROM users WHERE frac = {frac}"):
name, dostup = value
member = bot.get_user(name)
print(member)
await ctx.send(embed=discord.Embed(description=f" Пользователь {member.mention} | Доступ - {dostup}"))
else:
for value in sql.execute(f"SELECT idds, dostup FROM users WHERE frac = {frac} AND dostup = {dos}"):
name, dostup = value
member = bot.get_user(name)
print(member)
await ctx.send(embed=discord.Embed(description=f" Пользователь {member.mention} | Доступ - {dostup}"))```
How to output all lines in one embed
Ответы (1 шт):
Автор решения: Çℏėτᶄắɤẵ Çℏėτɤᶉᶄắ
→ Ссылка
Например, так:
sp=[]
for value in sql.execute(f"SELECT idds, dostup FROM users"):
name, dostup = value
member = bot.get_user(name)
print(member)
sp.append(f" Пользователь {member.mention} | Доступ - {dostup}")
a='\n'.join(sp)
await ctx.send(embed=discord.Embed(description=f'{a}'))
Выводиться будет в одном embed
