使用两个字典登录

Question

def login():

while True: 
    username = input("Please enter your username or x to quit: ")
    if username.casefold() == "x":
        break
    else:
        password = input("\nPlease enter your password: ")

    if username =="" or password =="":
        print("You cannot leave this blank,Please try again.")


    else:
        if len(username)>0 and len(password)>0: 
            for i in memberDict:
                if username == i and password == memberDict[i]:
                    print(f"Login successful, Welcome Member, {memberDict[i]}")
                    statuslist.append(0)
                    break
                else:
                    for i in adminDict:
                        if username == i and password == adminDict[i]:
                            print(f"Login successful, Welcome Admin, {adminDict[i]}")
                            statuslist.append(1)
                            break
                        else:
                            print("Login failed, There is no registered customers. Please try again.")

我试图通过两个不同的字典登录。字典工作正常，只是有问题，必须阅读他们两个和分开他们作为管理用户或成员

我正在尝试附加0作为成员，1作为管理员

Answer 1

第二个“admin”循环出现在成员dict循环中

也就是说，你应该：

for i in memberDict:
    if username == i and password == memberDict[i]:
        print(f"Login successful, Welcome Member, {memberDict[i]}")
        statuslist.append(0)
        login=True
for i in adminDict:
    if username == i and password == adminDict[i]:
        print(f"Login successful, Welcome Admin, {adminDict[i]}")
        statuslist.append(1)
        login=True
if not login:
    print("Login failed, There is no registered customers. Please try again.")

但事实上，你可能不应该重复这些格言。您可以通过以下方式判断键i是在dict memberDict还是adminDict中

if i in memberDict:
    if memberDict[i] == password:
       # Display login message

行政官也是如此

---

为清晰起见，请编辑：

我想说的是，如果dict中存在一对key和value的话，您正在尝试定位它。您不需要迭代dict（也就是说，使用for循环）。您可以使用if i in memberDict->；找出key是否在dict中（在您的例子中，是MemberDict中的用户名）；如果键i在dict中，则返回True，否则返回false

Answer 2

如果memberDict和adminDict的内容不重复，那么您确实需要检查这两个dict中的所有用户，以检查用户是否存在，只需使用找到的变量，这样就可以将时间复杂度从O（member）*O（admin）降低到O（member）+O（admin）

def login():
    while True: 
        username = input("Please enter your username or x to quit: ")
        if username.casefold() == "x":
            break
        else:
            password = input("\nPlease enter your password: ")

        if username =="" or password =="":
            print("You cannot leave this blank,Please try again.")


        else:
            if len(username)>0 and len(password)>0: 
                found = False
                for i in memberDict:
                    if username == i and password == memberDict[i]:
                        print(f"Login successful, Welcome Member, {memberDict[i]}")
                        statuslist.append(0)
                        found = True
                        break
                for i in adminDict:
                    if username == i and password == adminDict[i]:
                        print(f"Login successful, Welcome Admin, {adminDict[i]}")
                        statuslist.append(1)
                        found = True
                        break
                if not found:
                    print("Login failed, There is no registered customers. Please try again.")