Simple Python login Authentication applet

Use Python to write the login authentication applet, the user 3 times in a row password error is locked users.

############# Start ###############

#!/usr/bin/env python

Import OS

Import Sys

#

Os.system (' Clear ')

UserFile = File (' user_id.txt ', ' r+ ')

UserList = []

UserDict = {}

If Os.path.isfile ("User_id.txt"):

Pass

Else

print ' does not define user's files! ‘

Sys.exit

#userfile. ReadLines ()

For Userline in UserFile:

Useritem = Userline.strip ()

#生成系统的用户列表

Value_useritem = Useritem.split (';')

#基本判断添加取出

Value_username = Value_useritem[0].strip ()

VALUE_PASSWD = Value_useritem[1].strip ()

lock_value = Int (value_useritem[-1])

logincount_value = Int (value_useritem[-2])

#生成用户名列表

Userdict[value_username] = {' name ': Value_username, ' pwd ': value_passwd, ' lockcount ': logincount_value, ' locknum ': Lock _value}

Flag = ' Ture '

Counter = 0

While flag:

Username = raw_input (' Please enter user name: ')

USERPASSWD = raw_input (' Please enter password: ')

#判断是否是系统用户

If username not in Userdict.keys ():

print ' This user does not exist! ‘

Break

Elif userdict[username][' locknum '] = = 0 and userdict[username][' Lockcount '] < 3:

if userpasswd = = userdict[username][' pwd '].strip ():

print ' Welcome to landing system!!! ‘

Break

Else

Counter + = 1

userdict[username][' lockcount ') + = 1

UserFile = File (' user_id.txt ', ' w+ ')

For T in Userdict.values ():

Wuserlist = [Str (t[' name ']), str (t[' pwd "), str (t[' lockcount ']), str (t[' locknum '])

#wuserlist = T.values ()

Wuserlist_str = '; '. Join (Wuserlist)

#userfile. Seek (0)

Userfile.write (wuserlist_str + ' \ n ')

If counter >2:

print ' password entered error reached 3 times, exit login. ‘

Break

Else

print ' The user account has been locked! ‘

Sys.exit (' Please contact the administrator to unlock! ‘)

Continue

Userfile.close ()

############# End ###############

Where the format of the User_id.txt is:

User name, password, number of incorrect password, administrator to manually lock the account as follows:

[email protected] day1]# cat User_id.txt

Hejp;123;0;0

Test;456;0;0

User HEJP input three times password error, will display in User_id.txt error password number is 3, at this time the user locked, as long as it changed to 0 can be unlocked.

[email protected] day1]# cat User_id.txt

Hejp;123;3;0

Test;456;0;0

This article is from the "Early bird has the worm to eat" blog, please be sure to keep this source http://hejianping.blog.51cto.com/11279690/1783052

Simple Python login Authentication applet