在 python3 中从 SQLite3 中删除答案

【问题标题】：Deleting from SQLlite3 in python3在 python3 中从 SQLite3 中删除
【发布时间】：2021-11-08 09:00:48
【问题描述】：

我刚开始编码并尝试从 SQLite 数据库中删除用户 ID。我已经用谷歌搜索并观看了数小时的视频，试图找到一种方法让这段代码工作。请帮忙。我在树莓派上用 Python 3.X 编写，并使用 tkinter 作为 GUI。我的数据库有 4 列（first_name、last_name、user_id 和 password）。 password_delete 函数设置屏幕以获取输入，而 delete_funtion 应该删除输入的用户 id。目前我无法在两个函数之间传递 user_id 变量。一旦变量被传递，删除函数会起作用吗？非常感谢您的帮助。

from tkinter import *
import sqlite3

root = Tk()
root.geometry('500x700')

def delete_function():
    conn = sqlite3.connect('instructor_db.db')
    c = conn.cursor()
    
    user_id_delete = password_delete()
    print(user_id_delete)
    
    delete_variable = """DELETE FROM instructors WHERE user_id = ?"""
    c.execute(delete_variable,(user_id_delete, ))

    conn.commit()
    conn.close()

def password_delete():
    password_delete_screen = LabelFrame(root, width=500, height=700)
    password_delete_screen.pack()
    password_delete_screen.propagate(0)
    
    user_id_lbl = Label(password_delete_screen, text = "User Name").grid(row=0, column=0)
    
    global user_id_delete
    
    user_id_delete = Entry(password_delete_screen).grid(row=0, column=1)
    
    log_in_btn = Button(password_delete_screen, text = "Delete User", command = delete_function)
    log_in_btn.grid(row=2, column=0, columnspan=2)
    

password_delete()

root.mainloop()

【问题讨论】：

标签： python python-3.x sqlite tkinter

【解决方案1】：

首先，您在 SO 在以下行中多次询问了一个常见错误：

user_id_delete = Entry(password_delete_screen).grid(row=0, column=1)

user_id_delete 是 None 因为它是 grid(...) 的结果，而不是 Entry(...)。你需要把它分成两行：

user_id_delete = Entry(password_delete_screen)
user_id_delete.grid(row=0, column=1)

对于您的问题，您可以直接将输入内容传递给delete_function()，而不是使用全局变量：

def delete_function(user_id_delete):
    if user_id_delete:
        conn = sqlite3.connect('instructor_db.db')
        c = conn.cursor()

        delete_variable = """DELETE FROM instructors WHERE user_id = ?"""
        c.execute(delete_variable, (user_id_delete,))
        if c.rowcount:
            print('User deleted:', user_id_delete)
        else:
            print('User not found:', user_id_delete)

        conn.commit()
        conn.close()
    else:
        print('User input is empty')

def password_delete():
    password_delete_screen = LabelFrame(root, width=500, height=700)
    password_delete_screen.pack()
    password_delete_screen.propagate(0)

    Label(password_delete_screen, text = "User Name").grid(row=0, column=0)

    user_id_delete = Entry(password_delete_screen)
    user_id_delete.grid(row=0, column=1)

    log_in_btn = Button(password_delete_screen, text = "Delete User",
                        command=lambda: delete_function(user_id_delete.get()))
    log_in_btn.grid(row=2, column=0, columnspan=2)

【讨论】：