密码生成器有效密码检查

如何解决密码生成器有效密码检查

我正在用字典构建自己的密码生成器，并检查内部每种类型是否都有字符。它工作正常，但我认为我对支票进行了编码。

如果有一种更好的编码方法，您是否有想法？ 并且是否有一种方法可以摆脱检查，如果它已经准备就绪，那么它就不会检查其他类型？

PS：我想定义自己的二手低价/高档/特殊商品/数字，以便始终避免添加不喜欢的字符。

chars = ""
alpha_lowers = "abcdefghijklmnopqrstuvwxyz"
alpha_uppers = "ABCDEFGHIJKLMnopQRSTUVWXYZ"
specials = "$%&/()=?.,"
nums = "0123456789"
dictionary = {
    "a" : "anton","b" : "berta","c" : "caesar","d" : "dora","e" : "emil","f" : "friedich","g" : "gustav","h" : "hotel","i" : "india","j" : "julia","k" : "kilo","l" : "ludwig","m" : "marta","n" : "nordpol","o" : "otto","p" : "paula","q" : "quelle","r" : "richard","s" : "iegfried","t" : "theodor","u" : "ulrich","v" : "viktor","w" : "willhelm","x" : "xaver","y" : "ypsilon","z" : "zeppelin","A" : "Anton","B" : "Berta","C" : "Caesar","D" : "Dora","E" : "Emil","F" : "Friedrich","G" : "Golf","H" : "Hotel","I" : "India","J" : "Julius","K" : "Kilo","L" : "Ludwig","M" : "Marta","N" : "nordpol","O" : "otto","P" : "Paula","Q" : "Quelle","R" : "Richard","S" : "Siegfried","T" : "Theodor","U" : "Ulrich","V" : "Viktor","W" : "Willhelm","X" : "Xaver","Y" : "Ypsilon","Z" : "Zeppelin","$" : "Dollar","%" : "Prozent","&" : "Und","/" : "Schräg","(" : "Klammer auf",")" : "Klammer zu","=" : "Gleich","?" : "Fragezeichen","." : "Punkt","," : "Beistrich","0" : "Null","1" : "Eins","2" : "Zwei","3" : "Drei","4" : "Vier","5" : "Fünf","6" : "Sechs","7" : "Sieben","8" : "Acht","9" : "Neun"
}
all_chars = True

# Kleinbuchstaben hinzufügen // Adding Lowers
chars = chars + alpha_lowers

# Großbuchstaben hinzufügen // Adding uppers
chars = chars + alpha_uppers

# Spezial-Zeichen hinzufügen // Adding Specials
chars = chars + specials

# Nummern hinzufügen // Adding Nums
chars = chars + nums

# PW-Menge definieren // How many PW
password_n = 10

# PW-Länge definieren // Password length
password_len = 32


#--------------------------------------------------------------
def password_gen(length):

    # Generating PW
    password = ""
    for i in range (0,length):
        password = password + random.choice(chars)

    # Check if there is a Char from every type    
    if all_chars == True:
        in_alpha_lowers = False
        in_alpha_uppers = False
        in_specials = False
        in_nums = False
        for c in password:
            if in_alpha_lowers == False:
                if c in alpha_lowers:
                    in_alpha_lowers = True
            if in_alpha_uppers == False:
                if c in alpha_uppers:
                    in_alpha_uppers = True
            if in_specials == False:
                if c in specials:
                    in_specials = True
            if in_nums == False:
                if c in nums:
                    in_nums = True
        if in_alpha_lowers == False or in_alpha_uppers == False or in_specials == False or in_nums == False:
            print(password + " is not valid! New Passwort will be generated!" + "\n")
            return "invalid"
        else:        
            return password
    else:
        return password

#--------------------------------------------------------------
i = 1
while i <= password_n:
    password = ""
    sentence = ""
    password = password_gen(password_len)
    
    if password != "invalid":
        print("valid Passwort")
        i += 1
        for c in password:
                sentence = sentence + " " + dictionary[c] 

        print(password)
        print(sentence.lstrip() + "\n")

解决方法

您是否有想法，如果有一种更好的编码方式。

我建议看一下set的操作intersect。例如，您可以按照以下方式检查密码是否包含小写字母：

alpha_lowers = "abcdefghijklmnopqrstuvwxyz"
password = "password"
haslower = bool(set(password).intersection(alpha_lowers))
print(haslower)  # True

说明：我确实使alpha_lowers和password都通用一组字母，然后将其转换为bool，如果False为空且否则set。

是否有办法摆脱检查是否已经准备就绪，以便不检查其他类型？

由于您已经具有功能，可能在检查未通过后立即True。