为什么我一直出错，有人可以解释并帮助我吗

Question

我正在尝试制作一个简单的售票电影院，我正在复制这个特定的 YouTuber 作品以供参考，但我一直遇到错误，它说的是这个（看图片）。我不知道这里有什么不匹配，因为我已经检查了多次，这是代码：

import sqlite3

def create_table():
    conn = sqlite3.connect('Reservation.db')
    cursor = conn.cursor()

    cursor.execute('''
        CREATE TABLE IF NOT EXISTS Tickets (
            ticket_id INTEGER PRIMARY KEY,
            movie_name TEXT,
            ticket_quantity INTEGER,
            ticket_price INTEGER
        )''')

    conn.commit()
    conn.close()
    
def insert_Tickets():
    conn = sqlite3.connect('Reservation.db')
    cursor = conn.cursor()

    Tickets_data = [
        ('T1', 'Movie1', 3, 50),
        ('T2', 'Movie2', 2, 40),
        ('T3', 'Movie3', 4, 60),
        ('T4', 'Movie4', 5, 60),
        ('T5', 'Movie5', 1, 65)
    ]

    cursor.executemany('INSERT OR REPLACE INTO Tickets (ticket_id, movie_name, ticket_quantity, ticket_price) VALUES (?, ?, ?, ?)', Tickets_data)

    conn.commit()
    conn.close()


def get_tickets():
    conn = sqlite3.connect('Reservation.db')
    cursor = conn.cursor()
    cursor.execute('SELECT * FROM Tickets')
    tickets = cursor.fetchall()
    conn.close()
    
    return tickets

def update_quantity(id,reserved_quantity):
    conn = sqlite3.connect('Reservation.db')
    cursor = conn.cursor()
    cursor.execute('UPDATE Tickets SET ticket_quantity = ticket_quantity - ? WHERE ticket_id = ?',(reserved_quantity,id))
    
    conn.commit()
    conn.close()
    
create_table()
insert_Tickets()

Answer 1

如果为主键添加一列，它会起作用： 将您的创建表更改为：

cursor.execute('''
        CREATE TABLE IF NOT EXISTS Tickets (
            id INTEGER NOT NULL PRIMARY KEY,
            ticket_id Text,
            movie_name TEXT,
            ticket_quantity INTEGER,
            ticket_price INTEGER
        )''')