首页 > 编程 > Python > 正文

Python实现简单登录验证

2020-01-04 17:32:42
字体:
来源:转载
供稿:网友
这篇文章主要为大家详细介绍了Python实现简单登录验证的相关资料,需要的朋友可以参考下
 

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

#coding=utf-8__author__ = 'wangwc'import sys,oscount = 0locked = 0mark_user = 0mark_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/n")        else:          print ("Username or password wrong!")      continue

以上就是本文的全部内容,希望对大家的学习有所帮助。


发表评论 共有条评论
用户名: 密码:
验证码: 匿名发表