Question

我是Python的新手，正在尝试使用伪造的OTP系统和登录限制器创建简单的登录。 OTP有效，但登录限制器的计数器无效。仅一次尝试失败（我想要3次）后，它就会给我所需的结果。尝试失败后的输出为：

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错误的用户名或密码。请重试。

以退出代码1完成的过程

代码如下：

def old_acc():
count = 0  # count created to limit number of failed logins
login = input("Username:    ")  # prompts user to login in with their username
pw = input("Password:   ")  # prompts user to login in with their password
while count <= 3:
    for line in open("db.txt", "r").readlines():
        acc_info = line.split()

        # if username and pw do not match, prompt user to try again
        if login != acc_info[0] and pw != acc_info[1]:
            print("\nIncorrect Username or Password. Please try again.\n")
            count += 1
        # if username and pw match, login is successful; generate otp
        else:
            gen_otp()
            print("ACCESS GRANTED")
            access_info()

        # if failure count is = 3, deny access and lock out.
        if count == 3:
            # stops code and doesn't allow any further input.
            sys.exit("ACCESS LOCKED. YOU DON'T DESERVE TO SEE WHAT'S HERE. GOODBYE.")

以下是用于生成OTP的代码，供参考。

def gen_otp():
    digits = "0123456789"  # digits for OTP generation
    otp = ""

    for i in range(4):
        otp += digits[math.floor(random.random() * 10)]
    mbox("Enter OTP", otp, 1)  # gives user message with OTP
    otp_input = input("Enter OTP:   ")

    if otp == otp_input:
        print("ACCESS GRANTED")
        access_info()

    return otp

谢谢。

Answer 1

您的for循环检查db.txt文件中的每一行，如果与密码不匹配，则使计数器递增。假设db.txt可能包含多个密码，则该计数器在第一次尝试中将已经达到4。仅当db.txt的NO行与密码匹配时，才想增加计数器。

def old_acc():
   count = 0  # count created to limit number of failed logins
   success = False # keeps track of succesful login
   while count <= 3 and not success:
      login = input("Username:    ")  # prompts user to login in with their username
      pw = input("Password:   ")  # prompts user to login in with their password

       for line in open("db.txt", "r").readlines():
          acc_info = line.split()

           # if username and pw match, login is successful; generate otp
           if login == acc_info[0] and pw == acc_info[1]:
               gen_otp()
               print("ACCESS GRANTED")
               access_info()
               success = True
               break

        # if username and pw do not match, prompt user to try again
    if not success:
        print("\nIncorrect Username or Password. Please try again.\n")
        count += 1

    # if failure count is = 3, deny access and lock out.
    if count == 3:
        # stops code and doesn't allow any further input
        sys.exit("ACCESS LOCKED. YOU DON'T DESERVE TO SEE WHAT'S HERE. GOODBYE.")

Python登录计数器

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