for循环中嵌套if语句中的特定缩进错误

Question

我有以下代码，在登录功能中，输出错误（逻辑错误）。它基本上打印 "invalid username and password" 直到它到达正确的那个，然后打印 "correct login"。

错误输出：

例如，测试数据：user3和pass3，输出为：

*****LOGIN SCREEN******
Username: user3
Password: pass3
invalid username or password
invalid username or password
correct login
>>>

这里是有问题的代码，参考LOGIN函数：

usernames=["user1","user2","user3"]
passwords=["pass1","pass2","pass3"]

def main():
   mainmenu()


def mainmenu():
   print("****MAIN MENU****")
   print("=======Press L to login :")
   print("=======Press R to register :")
   choice1=input()
   if choice1=="L" or choice1=="l":
      login()
   elif choice1=="R" or choice1=="r":
      register()
   else:
      print("please make a valid selection")

usernames=["user1","user2","user3"]
passwords=["pass1","pass2","pass3"]

def login():
  print("*****LOGIN SCREEN******")
  username=input("Username: ")
  password=input("Password: ")
  for index_of_current_user, current_user in enumerate(usernames): #enumerate allows to you to go throw the list and gives to you the current element, and the index of the current element
    if username == current_user and passwords[index_of_current_user] == password: #since the two list are linked, you can use the index of the user to get the password in the passwords list
      print("correct login")
    else:
      print("invalid username or password")

def register():
  print("*****REGISTRATION****")
  username=input("Enter a username:")
  password=input("Enter a password:")
  users_pass[username] = password
  answer=input("Do you want to make another registration?")
  if answer=="y":
    register()
  else:
    registration_details()

def registration_details():
   print(usernames)
   print(passwords)

main()

我正在寻找 a) 解决方案并修复 问题，这样它只在找到正确的用户名和密码对时打印 "correct login" 一次，而不是循环打印每个 [=12] =]

b) 解释为什么这里的缩进或任何错误在这里不起作用 - 作为逻辑（在像 VB.Net 这样的语言中) 是健全的。

如果问题可以在没有额外代码的情况下解决（并且非常简单的修复），我会更喜欢它，但也许最优雅的解决方案是一个标志。也就是说，如果问题不仅仅是缩进或我错过的类似问题。

Answer 1

您之所以看到这种情况，是因为登录函数中的 else 语句。基本上你的代码当前在该函数中所做的是循环，检查用户名和密码是否等于当前值（即比较 user1 == user2）如果它们不相等，那么你会自动打印无效的用户名。

相反，您应该等到比较所有值后才打印无效的用户名或密码消息。此外 - 您可以添加一个中断以在找到有效值后停止 for 循环，而不是继续循环遍历您的值。您的登录功能将如下所示：

def login():
  print("*****LOGIN SCREEN******")
  username=input("Username: ")
  password=input("Password: ")
  found = False
  for index_of_current_user, current_user in enumerate(usernames): #enumerate allows to you to go throw the list and gives to you the current element, and the index of the current element
    if username == current_user and passwords[index_of_current_user] == password: #since the two list are linked, you can use the index of the user to get the password in the passwords list
      print("correct login")
      found = True
      break
  if found == False:
    print("invalid username or password")

这只会给您 1 个正确登录或无效用户名或密码的实例。

Answer 2

您可以利用异常以这种方式编写登录。

def login():
  print("*****LOGIN SCREEN******")
  username=input("Username: ")
  password=input("Password: ")
  try:
    index = usernames.index(username)
    if password == passwords[index]:
      print("correct login")
    else:
      print("invalid username or password")
  except:
    print("invalid username or password")

Answer 3

在您的 login() 函数中，您打印列表中的每个元素，因此您可以在循环后打印：

def login():
  print("*****LOGIN SCREEN******")
  username=input("Username: ")
  password=input("Password: ")
  correct_login = False
  for index_of_current_user, current_user in enumerate(usernames): #enumerate allows to you to go throw the list and gives to you the current element, and the index of the current element
    if username == current_user and passwords[index_of_current_user] == password: #since the two list are linked, you can use the index of the user to get the password in the passwords list
      correct_login = True
      break
  if(correct_login):
    print("correct login")
  else:
    print("invalid user name or password")

for循环中嵌套if语句中的特定缩进错误

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