Python 检查密码强度（如何检查条件）
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Checking the strength of a password (how to check conditions)
提问by user2412839
I am trying to create a system that requires you to enter a password. If it is all lower, upper or num then print weak, if it is two of the conditions, then it is med and if all have been met it is strong. It just does not seem to work.
我正在尝试创建一个需要您输入密码的系统。如果全部为lower、upper或num，则打印weak，如果是其中两个条件，则为med，如果全部满足则为strong。它似乎不起作用。
The weak and strong work however the medium does not.
弱和强的工作但是媒介没有。
I do not know where I have gone wrong.
我不知道我哪里出错了。
def password():
print ('enter password')
print ()
print ()
print ('the password must be at least 6, and no more than 12 characters long')
print ()
password = input ('type your password ....')
weak = 'weak'
med = 'medium'
strong = 'strong'
if len(password) >12:
print ('password is too long It must be between 6 and 12 characters')
elif len(password) <6:
print ('password is too short It must be between 6 and 12 characters')
elif len(password) >=6 and len(password) <= 12:
print ('password ok')
if password.lower()== password or password.upper()==password or password.isalnum()==password:
print ('password is', weak)
elif password.lower()== password and password.upper()==password or password.isalnum()==password:
print ('password is', med)
else:
password.lower()== password and password.upper()==password and password.isalnum()==password
print ('password is', strong)
回答by Martijn Pieters
password.isalnum()returns a boolean, so password.isalnum()==passwordwill alwaysbe False.
password.isalnum()返回一个布尔值，因此password.isalnum()==password将永远是False。
Just omit the ==passwordpart:
只是省略==password部分：
if password.lower()== password or password.upper()==password or password.isalnum():
# ...
Next, it can neverbe both all upper and lower, or all upper and numbers or all lower and all numbers, so the second condition (medium) is impossible. Perhaps you should look for the presence of someuppercase, lowercase and digits instead?
其次，它永远不可能既是高又是低，或者是全是高和数字，又是全是低和全数字，所以第二个条件（中）是不可能的。也许您应该寻找一些大写、小写和数字的存在？
However, first another problem to address. You are testing if the password is alphanumeric, consisting of just characters and/or numbers. If you want to test for just numbers, use .isdigit().
但是，首先要解决另一个问题。您正在测试密码是否为alphanumeric，仅由字符和/或数字组成。如果您只想测试数字，请使用.isdigit().
You may want to familiarize yourself with the string methods. There are handy .islower()and .isupper()methods available that you might want to try out, for example:
您可能希望熟悉字符串方法。有方便的.islower()和.isupper()可用的方法，你可能会想尝试，例如：
>>> 'abc'.islower()
True
>>> 'abc123'.islower()
True
>>> 'Abc123'.islower()
False
>>> 'ABC'.isupper()
True
>>> 'ABC123'.isupper()
True
>>> 'Abc123'.isupper()
False
These are faster and less verbose that using password.upper() == password, the following will test the same:
这些比 using 更快且更简洁，password.upper() == password以下将测试相同的内容：
if password.isupper() or password.islower() or password.isdigit():
# very weak indeed
Next trick you want to learn is to loop over a string, so you can test individual characters:
你想学习的下一个技巧是循环遍历一个字符串，这样你就可以测试单个字符：
>>> [c.isdigit() for c in 'abc123']
[False, False, False, True, True, True]
If you combine that with the any()function, you can test if there are somecharacters that are numbers:
如果将其与any()函数结合使用，则可以测试是否有一些字符是数字：
>>> any(c.isdigit() for c in 'abc123')
True
>>> any(c.isdigit() for c in 'abc')
False
I think you'll find those tricks handy when testing for password strengths.
我认为在测试密码强度时，您会发现这些技巧很方便。
回答by Inbar Rose
Here is a remake of what you wrote:
这是你写的内容的翻版：
import re
def password():
print ('Enter a password\n\nThe password must be between 6 and 12 characters.\n')
while True:
password = input('Password: ... ')
if 6 <= len(password) < 12:
break
print ('The password must be between 6 and 12 characters.\n')
password_scores = {0:'Horrible', 1:'Weak', 2:'Medium', 3:'Strong'}
password_strength = dict.fromkeys(['has_upper', 'has_lower', 'has_num'], False)
if re.search(r'[A-Z]', password):
password_strength['has_upper'] = True
if re.search(r'[a-z]', password):
password_strength['has_lower'] = True
if re.search(r'[0-9]', password):
password_strength['has_num'] = True
score = len([b for b in password_strength.values() if b])
print ('Password is %s' % password_scores[score])
Output (sample):
输出（样本）：
>>> password()
Enter a password
The password must be between 6 and 12 characters.
Password: ... ghgG234
Password is Strong
回答by milan
I was also looking for some password strength examine function, and found lot of half-worked suggestion. I assemblied my own function based on ones.
我也在寻找一些密码强度检查功能，并发现了很多不成功的建议。我根据自己的功能组装了自己的功能。
hope to help
希望有所帮助
def get_pw_strength( pw ):
s_lc = set(['a', 'c', 'b', 'e', 'd', 'g', 'f', 'i', 'h', 'k', 'j', 'm', 'l', 'o', 'n', 'q', 'p', 's', 'r', 'u', 't', 'w', 'v', 'y', 'x', 'z'])
s_uc = set(['A', 'C', 'B', 'E', 'D', 'G', 'F', 'I', 'H', 'K', 'J', 'M', 'L', 'O', 'N', 'Q', 'P', 'S', 'R', 'U', 'T', 'W', 'V', 'Y', 'X', 'Z'])
s_dg = set(['1', '0', '3', '2', '5', '4', '7', '6', '9', '8'])
s_sp = set(['+', ',', '.', '-', '?', ':', '_', '(', ')', '*', '/', ';', '+', '!'])
pw_s = 0
pw_steps = (5, 8, 12)
pw_l = len(pw)
if ( pw_l < 4 ):
return 0
for l in pw_steps :
if ( pw_l > l ):
pw_s += 1
#print "length over ", l," giving point", pw_s
c_lc = c_uc = c_dg = c_sp = 0
for c in pw :
if ( c in s_lc ) :
c_lc += 1
if ( c in s_uc ) :
c_uc += 1
if ( c in s_dg ) :
c_dg += 1
if ( c in s_sp ) :
c_sp += 1
if ( c_lc + c_uc + c_dg + c_sp <> pw_l ):
#print c_lc, c_uc, c_dg, c_sp, pw_l
#raise Exception "Forbidden chracter"
return -1
charset = 0
if ( c_lc ) :
pw_s += 1
charset = len(s_lc)
if ( c_uc ) :
pw_s += 1
charset = len(s_uc)
if ( c_dg ) :
pw_s += 1
charset = len(s_dg)
if ( c_sp ) :
pw_s += 2
charset = len(s_sp)
entropy = log(pow(charset,pw_l),2)
return pw_s, entropy
回答by ePi272314
Holá
The best approach is using regular expression search
Here is the function I am currently using
Holá
最好的方法是使用正则表达式搜索
这是我目前使用的函数
def password_check(password):
"""
Verify the strength of 'password'
Returns a dict indicating the wrong criteria
A password is considered strong if:
8 characters length or more
1 digit or more
1 symbol or more
1 uppercase letter or more
1 lowercase letter or more
"""
# calculating the length
length_error = len(password) < 8
# searching for digits
digit_error = re.search(r"\d", password) is None
# searching for uppercase
uppercase_error = re.search(r"[A-Z]", password) is None
# searching for lowercase
lowercase_error = re.search(r"[a-z]", password) is None
# searching for symbols
symbol_error = re.search(r"[ !#$%&'()*+,-./[\\]^_`{|}~"+r'"]', password) is None
# overall result
password_ok = not ( length_error or digit_error or uppercase_error or lowercase_error or symbol_error )
return {
'password_ok' : password_ok,
'length_error' : length_error,
'digit_error' : digit_error,
'uppercase_error' : uppercase_error,
'lowercase_error' : lowercase_error,
'symbol_error' : symbol_error,
}
EDIT:
Fallowing a suggestion of Lukasz here is an update to the especial symbol condition verification
编辑：
在这里放弃 Lukasz 的建议是对特殊符号条件验证的更新
symbol_error = re.search(r"\W", password) is None
