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}"))```

New line - new embed

How to output all lines in one embed


Ответы (1 шт):

Автор решения: Çℏėτᶄắɤẵ Çℏėτɤᶉᶄắ

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

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