任务是让用户输入密码，然后使用递归确保其中没有元音。如果是，则让用户重新输入密码。这就是我到目前为止所做的：
def passwordCheck(pwd):
"""checks if pwd has any vowels in it."""#doc string
vowels = 'aeiou'#specifies the characters that aren't allowed
if pwd == '':
return 0
elif pwd == None:
return None#Shouldn't be necessary but just in case
elif pwd[0] not in vowels:#checks that the 1st(0th) character is not a vowel
return passwordCheck(pwd[1:])#gets rid of the 1st(0th) character and starts again
elif pwd[0] in vowels:#checks if the 1st(0th) character is a vowel
return 1#if it is, stops the function calls and returns a value
password = str(input('Please enter a password with no vowels in it: '))#asks user to input their new password
x = passwordCheck(password)#checks the password is valid, i.e. no vowels
while x == 1:#when the password entered contains a vowel
print('\nSorry, that is not a valid password.\nYour password cannot contain any vowels.')#tells the user why their password is invalid
password = str(input('\nPlease enter a different password: '))#gives the user a chance to re-enter their password
x = passwordCheck(password)#checks to make sure the new password is valid
print('\nCongratulations, you have entered a valid password!')#tells the user if their desired password is valid
print('\nYou are now able to log on to the system with these credentials.')#could've been included on the previous line but looks neater here
我知道这可能不是最狡猾的方式，但在大多数情况下它适用于我。我希望听到更好的方式，但理想情况下，有人可以提供相同的风格。我不想在不理解它的情况下复制某些代码。
我的问题是处理用户根本没有输入密码的情况。第一个if语句：
if pwd == '':
return 0
我认为它刚刚处理了字符串完全递归后的情况，即没有元音，但经过几分钟检查后，显然这也适用于没有密码。 我也尝试过使用：
if pwd == None:
return something
现在我认为问题可能是因为我说：
password = str(input('######'))
但是我也在摆弄它，但仍然无法做到这一点！我试过google并搜索stackoverflow但没有运气，所以如果有人有任何想法/解决方案他们认为可能会有所帮助我会非常感谢听到他们。非常感谢你。
如何区分一个空的字符串，因为它已经被递归而用户没有输入任何内容？
最终使用
def passwordValid(pwd):
if len(pwd)>0 and passwordCheck(pwd)==0:
return pwd
else: return 'Fail'
password = str(input('Please enter a password with no vowels in it: '))#asks user to input their new password
y = passwordValid(password)#checks the password is valid, i.e. no vowels
while y == 'Fail':#when the password entered contains a vowel
print('\nSorry, that is not a valid password.\nYour password cannot contain any vowels or be empty.')#tells the user why their password is invalid
password = str(input('\nPlease enter a different password: '))#gives the user a chance to re-enter their password
y = passwordValid(password)#checks to make sure the new password is valid
print('\nCongratulations, you have entered a valid password!')#tells the user if their desired password is valid
print('\nYou are now able to log on to the system with these credentials.')#could've been included on the previous line but looks neater here
谢谢Wayne Werner修复标题和主要问题。
答案 0 :(得分：1)
这个问题可以分解为（至少）三个不同的子问题：
您的代码应该反映这种结构。因此，您可以使用以下功能布局：
def has_vowels(string):
if not string: # check for empty string
return False # empty strings never have vowels
# TODO we have a non-empty string at this point and can use recursion
def is_valid_password(string):
return len(string) > 0 and not has_vowels(string)
def request_password():
while True: # use an endless loop here, we don't won't to repeat
# the "input" statement. We could also change this to
# something like `for i in range(3)` to allow only a limited
# number of tries.
passwd = input('Please enter a password with no vowels in it: ')
# TODO check if input is valid, if yes, return, if no, print an error
答案 1 :(得分：0)
不要尝试用单一方法解决这两个问题。你有两个ditinct critera：没有元音;最小长度。
def isPasswordValid(pwd):
return len(pwd) > 4 and not passwordCheck(password)
x = isPasswordValid(password)
...
你可以通过添加另一个参数来解决这个问题，这个参数表明循环了多少个字符，但这很笨拙并且没有任何实际好处。
答案 2 :(得分：0)
您无法区分空字符串和空字符串。但是，您可以将变量设置为None，或者设置为“__no_string_entered_yet”之类的字符串。那就是说，我不明白为什么你需要，看看其他答案。
答案 3 :(得分：0)
我相信这会解决您的问题：
我选择不使用if/elif/else支持构造它，以便有效字符“落空”
def pwd_check(s):
vowels = 'aeiou'
if len(s) == 0: return False # This is only valid in the first iteration
if s[0] in vowels: return False
if len(s) == 1: return True # Success: a 1 character pwd with no vowels
return pwd_check(s[1:])
我考虑过检查以确保没有传递像' '这样的字符串，但我没有看到明确要求的。 pwd_check(password.strip())解决了这个问题。
答案 4 :(得分：0)
这就是我喜欢做的事 为了好玩，我添加了密码的最小和最大长度条件：
def passwordCheck(pwd,vowels = 'aeiou',minimum=5,maximum=12):
if pwd == '':
return 0,None,None
elif pwd[0] in vowels:
return -1,None,None
else:
y = passwordCheck(pwd[1:])[0]
if y==-1:
return -1,None,None
else:
return y + 1,minimum,maximum
mess = 'Please enter a password with no vowels in it: '
while True:
x,miin,maax = passwordCheckstr(input(mess))
if x==-1:
mess = ('\nSorry, that is not a valid password.\n'
'Your password cannot contain any vowels.\n'
'Please enter a different password: ')
elif x==0:
mess = ('\nSorry, you must enter a password.\n'
'Please do enter a password: ')
elif x<miin:
mess = ('\nSorry, the password must have at least %d characters.\n'
'The string you entered has %d characters.\n'
'Please, enter a new longer password: ' % (miin,x))
elif x>maax:
mess = ('\nSorry, the password must have at most %d characters.\n'
'The string you entered has %d characters.\n'
'Please, enter a new shorter password: ' % (maax,x))
else:
print ('\nCongratulations, you have entered a valid password!\n'
'\nYou are now able to log on to the system with these '
'credentials.')
break
另一种算法。
我不满意将这样的元组返回-1，None，None
def check_password(forbidden,minimum,maximum):
def passwordCheck(pwd,cnt=0,forbid = forbidden,
miin=minimum,maax = maximum):
# cnt is the number of preceding turns of recursion
# that have been already executed.
if pwd == '':
if cnt==0:
# cnt==0 means that it's the first turn of recursion
# since pwd is '', it means no entry has been done
return 0
elif cnt<miin:
return -3
elif cnt>maax:
return -2
elif pwd[0] in forbid:
return -1
else:
if cnt in (-3,-2,-1):
return cnt
else:
return passwordCheck( pwd[1:] , cnt+1 )
mess = 'Please enter a password with no vowels in it: '
while True:
x = str(raw_input(mess)).strip()
y = passwordCheck(x)
if y==0: # inexistent string
mess = ('\nSorry, you must enter a password.\n'
'Please do enter a password: ')
elif y==-1: # string contains a vowel
mess = ('\nSorry, that is not a valid password.\n'
'Your password cannot contain any vowels.\n'
'Please enter a different password: ')
elif y==-2: # string too long
mess = ('\nSorry, the password must have at most %d characters.\n'
'The string you entered has %d characters.\n'
'Please, enter a new shorter password: ' % (maximum,len(x)))
elif y==-3: # string too short
mess = ('\nSorry, the password must have at least %d characters.\n'
'The string you entered has %d characters.\n'
'Please, enter a new longer password: ' % (minimum,len(x)))
else: # success
print ('\nCongratulations, you have entered a valid password!\n'
'You are now able to log on to the system with these credentials.')
break
# EXECUTION
check_password('aeiou',5,12)