Python实现简单登录验证-Python教程-PHP中文网

Python实现简单登录验证

php中文网

发布： 2016-06-10 15:05:17

原创

1421人浏览过

本文实例为大家分享了简单的python登录验证，供大家参考，具体内容如下

编写登录接口

要求：1、输入用户名密码

　　　2、认证成功后显示欢迎信息

　　　3、输错三次后锁定

Smile企业费用管理系统源码1.0

一、源码特点企业费用管理系统，有权限分配，登陆验证，新增角色，发布公告等二、功能介绍1、js的兼容性有个地方不行（比如模块排序，那个时候也是雏鸟一只，写了一小撮，现在用jq应该好处理的吧，ie里面没问题，大家发挥吧）2、里面的菜单和对应菜单下面的目录项可以根据需求自己添加的，有对应模块3、可以根据自己设定的角色添加对应的访问页面4、有些操作涉及到按钮权限，对于这种思路，我粗粗的写了2个自定义控件，

查看详情

立即学习“Python免费学习笔记（深入）”；

#coding=utf-8
__author__ = 'wangwc'

import sys,os
count = 0
locked = 0
mark_user = 0
mark_passwd = 0
#获取路径
def cur_file_dir():
  path = sys.path[0]
  if os.path.isdir(path):
    return path
  elif os.path.isfile(path):
    return os.path.dirname(path)
#print (cur_file_dir())
path = cur_file_dir()
#print(path)
path1 = path.replace("\",'/') + '/'
#print (path1)
#path2 = path1 + '/'

#循环输入
while count < 3:
  name = input("Username:").strip()
  if len(name) == 0:
    print ("Username can not be empty....")
    continue
  key = input("Password:").strip()
  if len(key) == 0:
    print("The password can not be empty!Try again...")
    continue
  f = open(path1 + "username.txt","r")
  userlist = f.readlines()
  for user in userlist:
    if user.strip() == name:
      mark_user = 1
  f.close()

  if mark_user == 1:
    f = open(path1 + "%s_lock.txt" %(name),"r")
    locked = int(f.readline().strip())
    f.close()
  else:
    print ("Username or Passsord wrong....")
    break
  if locked == 1:
    print("Sorry, the username had been locked!!!Please call the system administrator...")
  else:
    f = open (path1 + "%s_passwd.txt" %(name),"r")
    passwd = (f.readline().strip())
    if passwd.strip() == key:
      mark_passwd = 1
    if mark_user == 1 and mark_passwd == 1:
      f = open("%s_count.txt" %(name),"w")
      f.write("0")
      f.close()
      print("%s,welcome BABY!" %(name) )
      #input('Press Enter to exit')
    else:
      f = open("%s_count.txt" %(name),"r")
      count = int((f.read().strip()))
      f.close()
      count +=1
      f = open("%s_count.txt" %(name),"w")
      f.write(str(count))
      f.close()
      print ("Username or password wrong!And the username '%s' has %d more chances to retry!" %(name,3 - count))
      if(count == 3):
        print ("'%s' has been locked!!!" %(name))
        if os.path.exists(path1 + "%s_lock.txt" %(name)):
          fobj = open(path1 + "%s_lock.txt" %(name),"w")
          fobj.writelines("1
")
        else:
          print ("Username or password wrong!")
      continue

登录后复制

以上就是本文的全部内容，希望对大家的学习有所帮助。