我可以改進當前的Python代碼嗎？

Question

我剛開始使用Python，並決定嘗試從Python Wiki這個小項目：

寫密碼猜測程序來跟蹤多少次用戶輸入的密碼錯誤。如果超過3次，則打印您被拒絕訪問。並終止程序。如果密碼正確，則打印您已成功登錄並終止程序。

這是我的代碼。它的作品，但它只是不覺得正確與這些環路中斷和嵌套如果語句。

# Password Guessing Program 
# Python 2.7 

count = 0 

while count < 3: 
    password = raw_input('Please enter a password: ') 
    if password != 'SecretPassword': 
     count = count + 1; 
     print 'You have entered invalid password %i times.' % (count) 
     if count == 3: 
      print 'Access Denied' 
      break 
    else: 
     print 'Access Granted' 
     break 

Answer 1

您可以用下面的函數替換while循環：

def login(): 
    for i in range(3): 
     password = raw_input('Please enter a password: ') 
     if password != 'SecretPassword': 
      print 'You have entered invalid password {0} times.'.format(i + 1) 
     else: 
      print 'Access Granted' 
      return True 
    print 'Access Denied' 
    return False 

您可能還需要考慮使用getpass模塊。

Answer 2

您可以將if語句帶出while循環。

# Password Guessing Program 
# Python 2.7 

count = 0 
access = False 

while count < 3 and not access: 
    password = raw_input('Please enter a password: ') 
    if password != 'SecretPassword': 
     count += 1 
     print 'You have entered invalid password %i times.' % (count) 
    else: 
     access = True 

if access: 
    print 'Access Granted' 
else: 
    print 'Access Denied' 

Answer 3

granted = False # default condition should be the least dangerous 
for count in range(3): 
    password = raw_input('Please enter a password: ') 
    if password == 'SecretPassword': # no need to test for wrong answer 
     granted = True 
     break 
    print 'You have entered invalid password %i times.' % (count+1) # else 

if granted: 
    print 'Access Granted' 
else: 
    print 'Access Denied' 

Answer 4

我不反對環路/如果「勢在必行」的感覺，但我會從你的「演示」分離的「商業邏輯」：

count = 0 

# Business logic 
# The correct password and the maximum number of tries is placed here 
DENIED, VALID, INVALID = range(3) 
def verifyPassword(userPassword): 
    global count 
    count += 1 

    if count > 3: 
     return DENIED 
    elif password == 'SecretPassword': 
     return VALID 

    return INVALID 

# Presentation 
# Here you do the IO with the user 
check = INVALID 
while (check == INVALID): 
    password = raw_input('Please enter a password: ') 
    check = verifyPassword(password) 

    if check == INVALID: 
     print 'You have entered invalid password %i times.' % (count) 
    elif check == VALID: 
     print 'Access Granted' 
    else # check == DENIED 
     print 'Access Denied'