Why is the table locked error?

Question

Python

I get this error when I simultaneously click on the inline button here is the code

if '21' not in mas_data_time:

                await bot.edit_message_reply_markup(call.message.chat.id, call.message.message_id)


                ass = 'Картой'
                conn_price_card = sqlite3.connect("Drive_in.db")  # или :memory: чтобы сохранить в RAM
                result_last_card = conn_price_card.execute(
                    f"""SELECT count FROM {user_id_start} WHERE id == 1""").fetchall()
                for back_elem_card in result_last_card:
                    mazafaka_card = back_elem_card[0]
                curr_Devit_pm.execute(f"UPDATE {user_id_start} SET oplata = '" + str(ass) + "' WHERE id =" + str(mazafaka_card))
                Deveti_pm = curr_Devit_pm.execute(f"""SELECT * FROM {user_id_start}""").fetchall()
                # Вывод результатов на экран
                count_back_rr = 1
                for Devet_pm in Deveti_pm:
                    await bot.send_message(call.from_user.id, str(count_back_rr) + ') ' + str(Devet_pm[7]))
                    abc = 1
                    table = 'Orders'
                    table_data = 'data_time'
                    a = '21'
                    curr_Devit_pm.execute(
                        f"""INSERT INTO {table} (name, user_name, shop, phone, price, bazar_tovar, oplata, clock) VALUES(?, ?, ?, ?, ?, ?, ?, ?)""",
                        (Devet_pm[3], Devet_pm[4], Devet_pm[5], Devet_pm[6], Devet_pm[7], Devet_pm[2], Devet_pm[11],
                         a))
                    curr_Devit_pm.execute(f"""DELETE FROM {user_id_start} WHERE id > 1""")
                    curr_Devit_pm.execute(f"UPDATE {user_id_start} SET count = '" + str(1) + "' WHERE id =" + str(1))
                    curr_Devit_pm.execute(f"UPDATE {user_id_start} SET location = '" + str(0) + "' WHERE id =" + str(1))
                    curr_Devit_pm.execute(f"""INSERT INTO {table_data} (clock) VALUES(?)""",
                                 (a,))
                con_Devit_pm.commit()
            else:
                await bot.send_message(call.message.chat.id, 'Это время уже забронированно выберите другое :(')

Answer 1

sqlite allows you to open only one connection to the database at a time, respectively, the first time you clicked - opened the database -> the database is blocked while it is open. Pressed a second time, sqlite tries to open it, but it is blocked because the first action did not complete.
Use more advanced tables for concurrent access