在python中使用多个变量创建列表是if语句的一部分

Question

   def runUser():
       x = raw_input("Enter your username:")
       y = raw_input("Enter your password :")
       users = [x == "aeduun" and y == "1234", x == "paul" and y == "fifty"]
       raw_input("")
       if x == users[] and y == users[]:#x == "aeduun" and y == "1234":
           print "you are now logged in"
       elif x == "Mercuryisle" and y == "shrek":
           print "Your account has expired..." \
                 "You will now bw taken back to the login page"
           time.sleep(5)
       return runUser()

下面是我作为python2.7小型测试项目的一部分编译的代码。在本文中，我试图用密码和用户名的变量列表来伪造一个基本登录名。该函数能够读取变量之间的关系，但包含对不同列表项的引用的错误。 当我启动程序时，只会调用一个无效的用户名（代码不显示）。 我想要一些关于如何初始化列表项的索引的建议，这些索引将作为if语句的一部分读取。你知道吗

Answer 1

def runUser():
   x = raw_input("Enter your username: ")
   y = raw_input("Enter your password: ")
   users = {"aeduun":"1234", "paul":"fifty"}
   raw_input("")
   if x in users:
       if y == users[x]:
           print "you are now logged in"
   elif x == "Mercuryisle" and y == "shrek":
       print "Your account has expired..." \
             "You will now bw taken back to the login page"
       time.sleep(5)
   return runUser()

运行方式：

>>> runUser()
Enter your username: aeduun
Enter your password: 1234

you are now logged in
Enter your username: Mercuryisle
Enter your password: shrek

Your account has expired...You will now bw taken back to the login page
Enter your username: 

还有一个建议，使用getpass.getpass()来获取密码，这样可以防止字符回音。你知道吗