主页 / 问题 / 1483974
Accepted
QuL1ux
Asked:2023-01-06 13:13:24 +0000 UTC2023-01-06 13:13:24 +0000 UTC

类型错误：元组索引必须是整数或切片，而不是 for 循环中的 str

  • 772

我试图创建一个领导团队( discord.py)，我已经有了一个包含用户的排序数组，但我无法使用其中的数据

数组示例：

{"702338551806951466": {"experience": 45, "level": 3, "need_xp": 70}}

代码本身：

@client.command(aliases=['l', 'лидеры'])
async def leaders(ctx):
    with open('users.json', 'r') as f:
      users = json.load(f)
    users = sorted(users.items(), key=lambda x: x[1]['level'], reverse=True)
    embed = discord.Embed(title='Лидеры 🏆', colour=discord.Colour.from_rgb(randint(5,255), randint(5,255), randint(5,255)))
    for user in users:
      level = user['level'] #ошибка тут
      embed.add_field(name=user.name, value=f'Уровень: **{level}**')
    await ctx.send(embed=embed)

错误：

TypeError: tuple indices must be integers or slices, not str

如何在循环中获取数组的元素？而如何获取数组的名字，也就是人的id呢？

python

2 个回答

  1. users = {"947336492487942174": {"experience": 6, "level": 3, "need_xp": 10},
         "702338551806951466": {"experience": 7, "level": 1, "need_xp": 10},
         "853231534433255270": {"experience": 8, "level": 4, "need_xp": 10}}
users = sorted(users.items(), key=lambda x: x[1]['level'], reverse=True)
# 1
for id_user, skill_user in users:
    print(f"{id_user} Уровень: ** {skill_user['level']} **")
# 2
for index, user in enumerate(users, 1):
    id_user, skill_user = user
    print(f"{index} {id_user} Уровень: ** {skill_user['level']} **")
    • 2
  2. Best Answer

    将level = user['level']标题改为user[1]['level']

    要获得id它，请联系user[0]

    上面，您按键和值对字典键值列表进行了排序level，因此users您现在有一个元组列表(<ключ>, <значение>)，示例元素在users：

    ('702338551806951466', {'experience': 45, 'level': 3, 'need_xp': 70})
    • 1

相关问题