16.python模拟登录程序和三级菜单

模拟登录程序需求：

1. 用户输入帐号密码进行登陆

2. 用户信息保存在文件内

3. 用户密码输入错误三次后锁定用户

#!/usr/bin/env python
#-*- coding: utf-8 -*-
i=0
while i<3:
    username = input("请输入您的用户名:")
    lock_file = open('account_lock','r')
    lock_list = lock_file.readlines()
    for lock_line in lock_list:
        lock_line =lock_line.strip('\n')
        if username == lock_line:
            lock_file.close()
            exit("您的账户已被锁定")
    user_account = open('user_account','r')
    user_list = user_account.readlines()
    for user in user_list:
        (user_infile,password_infile) = user.strip('\n').split()
        if username == user_infile:
            j=0
            while j<3:
                password=input("请输入您的密码:")
                if password == password_infile:
                    user_account.close()
                    exit("登录成功，欢迎您")
                else:
                    print("您输入的密码错误，您还有%s次机会输入"%(2-j))
                j += 1
            else:
                lock_file = open('account_lock','a')
                lock_file.write(username +'\n')
                exit("密码输入次数超限，您的用户已被锁定")
    else:
        print("您输入的用户名不存在，您还有%s次机会"%(2-i))
        i += 1
lock_file.close()
user_account.close()

三级菜单

作业需求：

1. 运行程序输出第一级菜单

2. 选择一级菜单某项，输出二级菜单，同理输出三级菜单

3. 菜单数据保存在文件中

4. 让用户选择是否要退出

5. 有返回上一级菜单的功能

 代码如下：

# Author:pengp
# !/usr/bin/env python
# -*- coding: utf-8 -*-
menu = {
    "上海": {
        "浦东": ["张江", "唐镇"],
        "嘉定": ["娄塘", "封浜"],
        "宝山": ["罗店", "月浦"]
    },
    "北京": {
        "海淀": ["上地", "中关村"],
        "朝阳": ["朝阳一镇", "朝阳二镇"]
    },
    "江苏": {
        "南京": ["浦口", "六合"],
        "苏州": ["昆山", "太仓"],
        "南通": ["如东", "启东"]
    },
    "湖北": {
        "武汉": ["汉口", "武昌"],
        "黄冈": ["麻城", "罗田"],
        "孝感": ["应城", "大悟"]
    }

}

current_level = menu
last_level = []
while True:
    for key in current_level:
        print(key)
    choice = input(">>>:")
    if len(choice) == 0:continue
    if choice == "b":
        if not last_level : break
        current_level = last_level[-1]
        last_level.pop()
    elif choice == 'q':
        exit("退出成功")
    if choice not in current_level:continue
    last_level.append(current_level)
    current_level = current_level[choice]