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python - 密码生成器有效密码检查

问题描述

我正在用字典构建自己的密码生成器，并检查里面是否有每种类型的字符。它工作正常，但我认为我对支票的编码有点复杂。

如果有办法以更好的方式对此进行编码，您是否有想法。如果它已经在较低的位置上，有没有办法摆脱检查，所以它不检查其他类型？

PS：我希望我自己定义使用的下限/上限/特价/数字，这样我就可以始终避免添加我不喜欢的字符。


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")

标签： pythonrandompasswordsgeneratorchange-password

解决方案

如果有办法以更好的方式对此进行编码，您是否有想法。

我建议看看它set的操作intersect。例如，您可以通过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：我确实将一组字母与alpha_lowersand共用password，然后将其转换为 bool，这导致Falseifset为空，True否则为空。

如果它已经在较低的位置上，有没有办法摆脱检查，所以它不检查其他类型？

由于您已经拥有功能，因此您可能会return "invalid"在检查未通过后立即进行操作。

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