項目需求描述：要求用戶輸入用戶名和密碼，認證成功後顯示歡迎信息，如果連續輸錯三次密碼則鎖定該用戶。
邏輯流程圖：
實現代碼：
#!/usr/bin/env python
import sys
account_file = 'account.txt'
lock_file = 'lock.txt'
# put accounts in a list
fh_account = open(account_file)
account_list = fh_account.readlines()
fh_account.close()
# every account has 3 chances of attempt at the beginning
retry_count = {}
for line in account_list:
line = line.split()
retry_count[line[0]] = 3
while True:
# put locked accounts in a list
fh_lock = open(lock_file)
lock_list = []
for i in fh_lock.readlines():
line = i.strip('\n')
lock_list.append(line)
fh_lock.close()
# handle the username and password empty issue
username = raw_input('Username: ').strip()
if len(username) == 0:
print '\033[31;1mUsername should not be empty !\033[0m'
continue
password = raw_input('Password: ').strip()
if len(password) == 0:
print '\033[31;1mPassword should not be empty !\033[0m'
continue
# authentication part
if username in lock_list:
print "\033[31;1mSorry, '%s' is locked already !\033[0m" % username
continue
if not retry_count.has_key(username): # inexistent account
retry_count[username] = 3
for line in account_list:
line = line.split()
if username == line[0] and password == line[1]: # authentication pass
retry_count[username] = 3 # reset retry times for this account
sys.exit('\033[32;1mWelcome %s login my system !\033[0m' % username)
else: # authentication failed
print '\033[31;1mWrong username or password !\033[0m'
retry_count[username] -= 1
if retry_count[username] > 0:
print '\033[31;1mYou have %d more chances !\033[0m' % retry_count[username]
else: # no more chance of attempt
fh = open(lock_file, 'a')
fh.write('%s\n' % username)
fh.close()
print "\033[31;1mSorry, '%s' is locked !\033[0m" % username