问题[sqlite3]

我用UPDATE、 、执行了不同的命令SELECT，INSERT并且不理解光标和 conn 之间的区别。两者的工作原理相同。你能解释一下吗

有一个员工表，其中有一个 fio 字段。文本。结果表中有字段sender_id 和receiver_id。两者都是整数。当尝试创建视图时

CREATE VIEW "problem_list_view" AS 
select problem_list.id,
        problem_list.date_start,
        users.fio as fio_sender,
        problem_list.problem_text,
        problem_list.solution_text,
        users.fio as fio_reciever,
        problem_list.date_end,
        statuses.name as status
from problem_list
left JOIN users on problem_list.sender_id = users.id
left join users on problem_list.receiever_id = users.id
left join statuses on problem_list.status_id = statuses.id

我得到错误模糊列名： And this is clear to me, sql indicates ambiguity. 我在结果表中添加了两个字段 fio_s 和 fio_r，这样就会有区别。

CREATE VIEW "problem_list_view" AS 
select problem_list.id,
        problem_list.date_start,
        users.fio_s as fio_sender,
        problem_list.problem_text,
        problem_list.solution_text,
        users.fio_r as fio_reciever,
        problem_list.date_end,
        statuses.name as status
from problem_list
left JOIN users on problem_list.sender_id = users.id
left join users on problem_list.receiever_id = users.id
left join statuses on problem_list.status_id = statuses.id

但我遇到了同样的错误。现在我不明白了。领域不同。告诉我哪里出了问题以及如何正确解决这个问题

sql3 = ("""SELECT password FROM inut_regist WHERE password =(?)""") cursor.execute(sql3, [user_password])

    sql4 = ('''SELECT login FROM inut_regist WHERE login =(?)''')
    cursor.execute(sql4, [user_login])
    
    if  :
        self.label.setText('Ошибка авторизации!')
    else:
        self.label.setText('Успешная авторизация!')

我在不和谐中有自己的机器人，如果用户不在数据库中，那么他必须在那里输入，但我无法在 on_message 事件中弄清楚如何做到这一点。

@bot.event
async def on_message(message):
    if cursor.execute(f"SELECT id FROM users WHERE id = {message.author.id}").fetchone()[0] is None:
        cursor.execute(f"INSERT INTO users VALUES ({message.author.id}, 100, 'Нету', 0, 0)")
        conn.commit()
        cursor.execute(f"INSERT INTO inventory VALUES ({message.guild.id}, {message.author.id}, '')")


    else:
        pass
    bot.process_commands(message)

检查用户是否有 id 时出错。但是如果它抛出错误，我如何检查它是否为 None

Traceback (most recent call last):
  File "C:\Users\semnf\AppData\Local\Programs\Python\Python310\lib\site-packages\discord\client.py", line 343, in _run_event
    await coro(*args, **kwargs)
  File "c:\Users\semnf\OneDrive\Рабочий стол\DiscBot\hello.py", line 62, in on_message
    if cursor.execute(f"SELECT id FROM users WHERE id = {message.author.id}").fetchone()[0] is None:
TypeError: 'NoneType' object is not subscriptable

这是我的代码：

import discord
from discord.ext import commands

import sqlite3
from config import settings

client = commands.Bot(command_prefix = settings['PREFIX'])
client.remove_command('help')

connection = sqlite3.connect('server.db')
cursor = connection.cursor()

@client.event
async def on_ready():
    cursor.execute("""CREATE TABLE IF NOT EXISTS users (
        name TEXT, 
        id INT,
        cash BIGINT,
        rep INT,
        lvl INT
    )""")
    connection.commit()

    for guild in client.guilds:
        for member in guild.members:
            cursor.execute(f"SELECT id FROM users where id={member.id}")
            if cursor.fetchone() is None:
                cursor.execute(f"INSERT INTO users (name, id, cash, rep, lvl) VALUES ('{member}', {member.id}, 0, 0, 1)")
                connection.commit()
            else:
                pass

print('Бот готов к работе!')

问题是只有创建的机器人被输入到我的数据库中。 其他参与者没有进入，尽管他们似乎是。帮忙求求！！