密码生成器 - 嵌套 while 循环

Question

我正在研究下面的密码生成器和一个让我困惑的问题。有一个嵌套的 while 循环询问用户是否愿意在生成密码后选择一个新密码。每次您选择“是”时，理想情况下它应该生成一个新密码来替换旧密码。事实并非如此，每次重新选择时，它都会在旧密码上添加一个新密码。我不在这里做什么？import randomfrom string import ascii_lowercase, ascii_uppercase, digitsspecial_chars = "!#$%&*@^"available_chars = list(ascii_lowercase) + list(ascii_uppercase) + list(digits) + list(special_chars)def get_length():    user_l = ""    while not user_l.isdigit() or int(user_l) < 6:        user_l = input("Please input a password length.\n")    return int(user_l)    print(int(user_l))def pwd_gen(length):    return [random.choice(available_chars) for i in range(length)]def pwd_chk(length):    pwd = []    while True:        pwd = pwd_gen(length)        if set(pwd) & set(ascii_lowercase) == set():            continue        elif set(pwd) & set(ascii_uppercase) == set():            continue        elif set(pwd) & set(digits) == set():            continue        elif set(pwd) & set(special_chars) == set():            continue        else:            print("\nYour password is " + "".join(pwd))            while True:                accept = input("Would you like a different password? Y/N\n\n")                if accept.lower() == "n" or accept.lower() == "no":                    print("\nYour password is:")                    break                elif accept.lower() == "y" or accept.lower() == "yes":                    pwd_chk(length)                    break                else:                    print("\nInvalid input. Your password is " + "".join(pwd))                    continue            break    pwd = "".join(pwd)    print(pwd)pwd_chk(get_length())

Answer 1

选择“Y”后并没有停止循环，因此未完成的函数将继续执行。通过使用return语句（而不是break），您可以停止函数打印它生成的密码。

import random

from string import ascii_lowercase, ascii_uppercase, digits

special_chars = "!#$%&*@^"

available_chars = list(ascii_lowercase) + list(ascii_uppercase) + list(digits) + list(special_chars)

def get_length():

    user_l = ""

    while not user_l.isdigit() or int(user_l) < 6:

        user_l = input("Please input a password length.\n")

    return int(user_l)

    print(int(user_l))

def pwd_gen(length):

    return [random.choice(available_chars) for i in range(length)]

def pwd_chk(length):

    pwd = []

    while True:

        pwd = pwd_gen(length)

        if set(pwd) & set(ascii_lowercase) == set():

            continue

        elif set(pwd) & set(ascii_uppercase) == set():

            continue

        elif set(pwd) & set(digits) == set():

            continue

        elif set(pwd) & set(special_chars) == set():

            continue

        else:

            print("\nYour password is " + "".join(pwd))

            while True:

                accept = input("Would you like a different password? Y/N\n\n")

                if accept.lower() == "n" or accept.lower() == "no":

                    print("\nYour password is:")

                    break

                elif accept.lower() == "y" or accept.lower() == "yes":

                    pwd_chk(length)

                    return # CHANGE THIS FROM "break" TO "return"! #

                else:

                    print("\nInvalid input. Your password is " + "".join(pwd))

                    continue

            break

    pwd = "".join(pwd)

    print(pwd)

pwd_chk(get_length())

作为旁注，如果您实际使用它们，则不应使用默认的随机模块来生成密码。使用该secrets模块或使用随机数生成器播种更安全

import os

import random

random.seed(os.urandom(16))

实现安全。

Answer 2

您可以添加一个标志而不是再次调用该函数（内部 while 代码）：

            while True:

                replace_password = False # Flag added here

                accept = input("Would you like a different password? Y/N\n\n")

                if accept.lower() == "n" or accept.lower() == "no":

                    print("\nYour password is:")

                    break

                elif accept.lower() == "y" or accept.lower() == "yes":

                    replace_password = True # Setting flag

                    break

                else:

                    print("\nInvalid input. Your password is " + "".join(pwd))

                    continue

            if replace_password: # Action based on flag

                continue

            break