检查密码强度(如何检查条件) [英] Checking the strength of a password (how to check conditions)

问题描述

我正在尝试创建一个要求您输入密码的系统.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.

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.

弱者和强者有效，但媒介无效.

The weak and strong work however the medium does not.

我不知道我哪里出错了.

I do not know where I have gone wrong.

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)

推荐答案

Holá
最好的方法是使用正则表达式搜索
这是我目前使用的功能

Holá
The best approach is using regular expression search
Here is the function I am currently using

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,
    }

在这里放弃 Lukasz 的建议是对特殊符号条件验证的更新

Fallowing a suggestion of Lukasz here is an update to the especial symbol condition verification

symbol_error = re.search(r"\W", password) is None

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