无法通过输入菜单访问代码内的功能

Question

我无法访问输入菜单中其余代码中的功能 用户应该输入一个字母，该字母将带他到程序的函数并继续按函数状态进行编辑，但我收到一条错误消息，指出“NameError：名称'reg_user'未定义”。

def display_menu_options():    

    global menu_options

    menu_options = input('''Please select one of the following options:

a - Adding a task
va - View all tasks
vm - View my tasks
e - Exit
:''').lower()

    if menu_options == "a":
        add_task()
    elif menu_options == "va":
        view_all()
    elif menu_options == "vm":
        view_mine()
    elif menu_options == "e":
        print("Goodbye!")
    else:
        print("You have made a wrong choice, Please Try again")

    return menu_options

def display_admin_menu_options():

    global menu_options

    menu_options = input('''Please enter one of the following options:
    
r - register user
a - add task
va- view all tasks
vm - view my tasks
gr - generate reports
ds - display statistics
e - exit
:''').lower()

    if menu_options == "r":
        reg_user()
    elif menu_options == "a":
        add_task()
    elif menu_options == "va":
        view_all()
    elif menu_options == "vm":
        view_mine()
    elif menu_options == "gr":
        generate_reports()
    elif menu_options == "ds":
        all_statistics()
    elif menu_options == "e":
        print("Goodbye!")
    else:
        print("You have made a wrong choice, Please Try again")

    return menu_options

def login():

    username = input("Enter your username: ")
    password = input("Enter your password: ")

    for line in open('user.txt', 'r').readlines():
        field = line.strip().split(", ")
        if username == field[0] and password == field[1]:
            print('Welcome' + username)
            return True, field[0] == "admin"

    return False, False

login_success, is_admin = login()

if login_success and is_admin:
    display_admin_menu_options()
elif login_success:
    display_menu_options()
else:
    print("Username or password incorrect!")

Answer 1

我假设您正在执行并将其编写为脚本。因此，您正在调用该函数，并在定义您要访问的函数的下方。这些不是“定义的”，因为代码是解释的而不是编译的。将所有函数定义放在 login() 调用之上