python tkinter与Mysql交互
IronSimon 人气:0本例已经实现的数据库password，数据库的表以及表结构如下：
表中已经插入的信息：
实现思路无非是用户完成账户密码输入并点击登录按钮后，程序先进行数据库连接，然后根据用户提供的参数，
发出相应的查询语句，根据返回的查询结果给出相应的响应。
代码实现
# -*- coding: utf-8 -*-
"""
Created on Tue Nov 6 14:29:54 2018
Description:实现tkinter的密码验证
1.与数据库验证
Version:
@author: HJY
"""
import tkinter as tk
from tkinter import messagebox
import sys
import pymysql
class loginf():
def __init__(self,master):
self.master = master
self.face = tk.Frame(self.master,)
self.face.pack()
tk.Label(self.face,text='账户').pack()
self.t_account = tk.Entry(self.face,)
self.t_account.pack()
tk.Label(self.face,text='密码').pack()
self.t_password = tk.Entry(self.face,)
self.t_password.pack()
btn_login = tk.Button(self.face,text='login',command=self.login)
btn_login.pack()
def login(self,):
account = self.t_account.get()
password = self.t_password.get()
#判空操作：略
print(account,password)
#数据库处理
connection = pymysql.connect(host='localhost',user='root',port=3306)
try:
with connection.cursor() as cursor:
command1 = "use password;"
command2 = "select password from passbook where account = (%s);"
cursor.execute(command1)
result = cursor.execute(command2,(account))
connection.close()
except:
sys.exit()
else:
if result == 0:
print('no this account!')
tk.messagebox.showerror('Info',"Account Not Exist！")
else:
print('查找结果：',result)
if cursor.fetchone()[0] == password:
print('Login successfully！')
tk.messagebox.showinfo('Info',"Login successfully！")
#销毁登陆界面，生成登陆后界面
self.face.destroy()
homef(self.master)
else:
print('password input error')
tk.messagebox.showerror('Info',"Password Error！")
class homef():
def __init__(self,master):
self.master = master
self.face = tk.Frame(self.master,)
self.face.pack()
btn_showinfo = tk.Button(self.face,text='info',command=self.showinfo)
btn_showinfo.pack()
def showinfo(self,):
pass
if __name__ == '__main__':
root = tk.Tk()
root.title('Login with password')
root.geometry('200x200')
loginf(root)
root.mainloop()
效果示例：
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