Python实现简单登录验证

#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