str.isdigit（）似乎沒有在python中工作

Question

我正在研究一個密碼檢查器，檢查字符串是否是有效的密碼。我必須檢查是否至少有八個字符，必須只包含字母和數字，最後兩個字符必須是數字。

除password.isdigit()以外，這一切似乎工作。有時密碼出來是有效的，有時它不會。有什麼建議麼？

# Gets the users password 
password = input('Enter a string for password: ') 
# Splices the last two characters of the password 
lastTwo = password[-2:] 

# Checks the password if it is less than 8 characters 
while len(password) < 8: 
    print('The password you entered is too short.') 
    print() 
    password = input('Enter a string for password: ') 

    # Checks the password if it is composed of letters and numbers 
    while password.isalnum() == False: 
     print('Your password has special characters not allowed.') 
     print() 
     password = input('Enter a string for password: ') 

    # Checks the spice to verify they are digits 
    while lastTwo.isdigit() == False: 
     print('Your last two characters of your password must be digits.') 
     print() 
     password = input('Enter a string for password: ') 

print('Your password is valid.') 

Answer 1

有與你提供的代碼問題了一把。特別是，您只能檢查後續規則while len(password) < 8。如果你給它一個長度爲10的密碼，規則從不被檢查。此外，你不更新lastTwo隨着每一個新的密碼嘗試解決這一問題將與if...elif..elif...else...包裹在一個整體while語句來代替你的幾個while聲明

的一種方法，具體如下：

# Gets the users password 
password = input('Enter a string for password: ') 

while True: 
    # Checks the password if it is less than 8 characters 
    if len(password) < 8: 
     print('The password you entered is too short.') 
    # Checks the password if it is composed of letters and numbers 
    elif not password.isalnum(): 
     print('Your password has special characters not allowed.') 
    # Checks the spice to verify they are digits 
    elif not password[:-2].isdigit(): 
     print('Your last two characters of your password must be digits.') 
    else: 
     # we only get here when all rules are True 
     break 

    print() 
    password = input('Enter a string for password: ') 

print('Your password is valid.') 

這應該按照你的意圖工作。但是，雖然我們在這，它爲什麼不告訴用戶每規則他們的密碼已損壞？從UI的角度來看，它有助於讓用戶瞭解情況。

如果我們存儲旁邊的相關規則是否被滿足的信息消息，我們可以快速計算出所有這一切都被打破，像這樣的規則：

valid_password = False 

while not valid_password: 
    # Get a password 
    password = input('\nEnter a string for password: ') 
    # applies all checks 
    checks = { 
     '- end in two digits': password[-2].isdigit(), 
     '- not contain any special characters': password.isalnum(), 
     '- be over 8 characters long': len(password) > 8 
    } 
    # if all values in the dictionary are true, the password is valid. 
    if all(checks.values()): 
     valid_password = True 
    # otherwise, return the rules violated 
    else: 
     print('This password is not valid. Passwords must:\n{}'.format(
      '\n'.join([k for k, v in checks.items() if not v]))) 

print('Your password is valid.') 

Answer 2

你永遠不會在你的while循環中更新你的lastTwo的值。因此，想象一下，如果用戶第一次輸入密碼abc123。那麼lastTwo將計算爲23。

現在您的代碼會發現密碼太短，並提示用戶輸入新密碼。假設他輸入abcdefgh。這現在通過你的第一次和第二次檢查。但請注意，lastTwo仍然是23，因此您的第三張支票將錯誤地通過。

你應該這樣重新計算，只要你接受一個新的密碼，或直接檢查這樣lastTwo的價值：

while (password[-2:]).isdigit() == False: