为什么用户仍然可以在密码中输入umgc，而不会返回为false

Question

下面是我的代码，当运行它并输入umgc时，它将接受密码。有人能帮上忙吗？非常感谢！

def strong_password(password):
Symbol=['#']
umgc=['umgc', 'umgC', 'umGc', 'umGC', 'uMgc', 'uMgC', 'uMGc', 'uMGC', 'Umgc', 'UmgC', 'UmGc', 'UmGC', 'UMgc', 'UMgC', 'UMGc', 'UMGC']
returned=True
if len(password) <= 6:
print('the length of password should be at least 6 characters long')
returned=False
if len(password) >= 12:
print('the length of password should be not be greater than 12')
returned=False
if not any(char in Symbol for char in password):
print('the password should have the # symbol')
returned=False
if password[0] == '#':
print('The # symbol should not be in the beginning of the password')
returned=False
if password[-1] == '#':
print('The password must not end with the # symbol')
returned=False
if any(char in umgc for char in password):
print('Password must not contain "UMGC" in any combination of upper/lowercase letters')
returned=False
if returned:
print('Strong Password!')
return returned

Answer 1

使用一个集合进行成员资格测试。

更改

if any(char in umgc for char in password):

至

umgc = set('umgc')
if umgc.issubset(password.lower()):

---

由于您询问单词列表中是否有单个字符，您的原始操作失败。你的原件是这样做的：

>>> umgc=['umgc', 'umgC', 'umGc', 'umGC', 'uMgc', 'uMgC', 'uMGc', 'uMGC', 'Umgc', 'UmgC', 'UmGc', 'UmGC', 'UMgc', 'UMgC', 'UMGc', 'UMGC']
>>> password = 'Umgc'     
>>> for char in password:
...     print(f'is {char} in {umgc}') 
...
is U in ['umgc', 'umgC', 'umGc', 'umGC', 'uMgc', 'uMgC', 'uMGc', 'uMGC', 'Umgc', 'UmgC', 'UmGc', 'UmGC', 'UMgc', 'UMgC', 'UMGc', 'UMGC']
is m in ['umgc', 'umgC', 'umGc', 'umGC', 'uMgc', 'uMgC', 'uMGc', 'uMGC', 'Umgc', 'UmgC', 'UmGc', 'UmGC', 'UMgc', 'UMgC', 'UMGc', 'UMGC']
is g in ['umgc', 'umgC', 'umGc', 'umGC', 'uMgc', 'uMgC', 'uMGc', 'uMGC', 'Umgc', 'UmgC', 'UmGc', 'UmGC', 'UMgc', 'UMgC', 'UMGc', 'UMGC']
is c in ['umgc', 'umgC', 'umGc', 'umGC', 'uMgc', 'uMgC', 'uMGc', 'uMGC', 'Umgc', 'UmgC', 'UmGc', 'UmGC', 'UMgc', 'UMgC', 'UMGc', 'UMGC']

Answer 2

输入"umgc"将接受密码，因为在这一行中：

if any(char in umgc for char in password):

您正在询问单词列表umgc:中的密码(word(中是否有字符

然而，这将起作用：

if any(char in words for words in umgc for char in password):

这里我们要问的是，密码中是否有任何字符与单词列表中任何单词的任何字符相匹配。

更好的解决方案已经在答案中
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