Question

不，我不是在诅咒标题。我需要创建一个密码处理程序来检查输入是否符合某个条件，其中一个条件是它必须包含一个字符$ @！％＆amp; * _。这就是我现在所拥有的。

def pword():
    global password
    global lower
    global upper
    global integer

password = input("Please enter your password ")
length = len(password)
lower = sum([int(c.islower()) for c in password])
upper = sum([int(c.isupper()) for c in password])
integer = sum([int(c.isdigit()) for c in password])


def length():
    global password
if len(password) < 8:
    print("Your password is too short, please try again")
elif len(password) > 24:
    print("Your password is too long, please try again")

def strength():
    global lower
    global upper
    global integer
if (lower) < 2:
    print("Please use a mixed case password with lower case letters")
elif (upper) < 2:
    print("Please use a mixed case password with UPPER case letters")
elif (integer) < 2:
    print("Please try adding numbers")
else:
    print("Strength Assessed - Your password is ok")

Answer 1

这种事情会起作用：

required='$@!%&*_'

def has_required(input):
    for char in required:
        if input.contains(char):
            return True
    return False

has_required('Foo')

Answer 2

must_have = '$@!%&*_'
if not any(c in must_have for c in password):
    print("Please try adding %s." % must_have)

如果any(c in must_have for c in password)中的任何一个字符也在password中，

must_have将返回True，换句话说，它将返回True，密码是好的。因为要测试错误的密码，我们将not放在它前面以反转测试。因此，此处仅针对错误密码执行print语句。

Answer 3

您可以使用列表推导+内置any()轻松实现此目的。

has_symbol = any([symbol in password for symbol in list('$@!%&*_')])

或者更多分手：

required_symbols = list('$@!%&*_')
has_symbol = any([symbol in password for symbol in required_symbols])

检查字符$ @！％＆amp; * _ python的输入

3 个答案: