python如何实现不用装饰器实现登陆器小程序

这篇文章主要介绍了python如何实现不用装饰器实现登陆器小程序,文中通过示例代码介绍的非常详细，对大家的学习或者工作具有一定的参考学习价值,需要的朋友可以参考下

下面代码演示了不使用装饰器实现用户登陆功能的小程序,在python3.x下可正常运行

#!/usr/bin/env python3
# -*- coding: utf-8 -*-
"""
Created on Tue Nov 26 15:26:58 2019

@author: root
"""

def menu():
 print('choice'.center(50,'-'))
 msg = '''
 1. home
 2. finance
 3. book
 4. exit
 '''
 print(msg)
 print("".center(50,'-'))

def login(login_flat,choice):
 if not login_flat:
   while True:
     username = str(input("username:"))
     password = str(input("password:"))
     if choice == '1':
       home()
       with open('jingdong.txt','r') as f1:
         for line in f1:
           element = line.split(" ",1)
           print(element[0])
           print(element[1])
           if username.strip() == element[0].strip() and password.strip() == element[1].strip():
             print("log in successful!")
             return True
           print("failed to log in!,try it again!")
     elif choice == '2':
       finance()
       with open('weixin.txt','r') as f2:
         for line in f1:
           element = line.split(" ",1)
           if username.strip() == element[0].strip() and password.strip() == element[1].strip():
             print("log in successful!")
             return True
           print("failed to log in!,try it again!")  
     elif choice == '3':
       book()
       with open('jingdong.txt','r') as f3:
         for line in f3:
           element = line.split(" ",1)
           if username.strip() == element[0].strip() and password.strip() == element[1].strip():
             print("log in successful!")
             return True
           print("failed to log in!,try it again!")              
 else:
   print("You have logged in before!")
   return True
#@login(login_flat,choice)            
def home():
 print("home.....")
#@login(login_flat,choice)
def finance():
 print("finance.....")
#@login(login_flat,choice)
def book():
 print("book.....")

login_flag = False
while True:
 menu()
 choice = input("your choice:")
 if choice == '1':
   login_flag = login(login_flag,choice)
 elif choice == '2':
   login_flag = login(login_flag,choice)
 elif choice == '3':
   login_flag = login(login_flag,choice)
 elif choice == '4':
   print('bye-bye')
   break
 else:
   print("wrong input ,try it again!")

来源：https://www.cnblogs.com/iceberg710815/p/11936947.html