检查密码的强度(如何检查条件)

use*_*839 9 python python-3.x

我正在尝试创建一个要求您输入密码的系统.如果它全部较低,则较高或数字则打印较弱,如果是两个条件,那么它是med,如果所有条件都满足则很强.它似乎没有用.

然而,媒介没有弱势和强势的工作.

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)
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ePi*_*314 23

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,
    }
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编辑:
在这里提出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
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  • 至少前导和尾随空格应该出错:`o.whitespace_error = len(pwd.strip()) != len(pwd)` (2认同)

Mar*_*ers 5

password.isalnum()返回一个布尔值,因此password.isalnum()==password始终False

只需省略==password部分:

if password.lower()== password or password.upper()==password or password.isalnum():
    # ...
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其次,它永远不能同时是所有上限和下限,也不能是全部上限和下限,也不能全部都是下限和下限,因此第二个条件(中)是不可能的。也许您应该寻找一些大写,小写和数字的出现?

但是,首先要解决的另一个问题。您正在测试密码是否为字母数字,仅由字符和/或数字组成。如果只想测试数字,请使用.isdigit()

您可能需要熟悉字符串方法。有方便的.islower().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
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与使用相比password.upper() == password,这些方法更快捷,更省力,以下将对其进行测试:

if password.isupper() or password.islower() or password.isdigit():
    # very weak indeed
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您要学习的下一个技巧是遍历字符串,以便可以测试各个字符:

>>> [c.isdigit() for c in 'abc123']
[False, False, False, True, True, True]
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如果将其与any()函数结合使用,则可以测试是否有一些数字字符:

>>> any(c.isdigit() for c in 'abc123')
True
>>> any(c.isdigit() for c in 'abc')
False
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我认为您在测试密码强度时会发现这些技巧很方便。