Python Curses：返回上一菜单

Question

我在使用ncurses做一个菜单系统。选择一个选项后会进入另一个菜单。但是，我该怎么返回呢？

import curses

def Main():
   x = 0
   while x!= ord('2'):
       x = screen.getch()
       screen.clear();screen.border();
       screen.addstr(1,1, "Please choose:")
       screen.addstr(3,1, "1 - Another Menu")
       screen.addstr(4,1, "2 - Exit")

       if x==ord('1'):
           y = 0
           while y!= ord('2'):
              y = screen.getch()
              screen.clear();screen.border();
              screen.addstr(1,1, "Please choose from new menu:")
              screen.addstr(3,1, "1 - Do Something new")
              screen.addstr(4,1, "2 - Previous Menu")
              if y == ord('1'): doSomething()

           #Here I exit the internal loop. I need to go back to the previous menu, but I don't know how.
           ##
   ##exit outside loop and close program
   ##
   curses.endwin(); exit();

screen = curses.initscr()
Main()

理想情况下，我需要使用GOTO模块在代码行之间跳转，但我用的设备没有这个模块。

你们知道其他方法吗？非常感谢任何帮助。

============ 更新: ==================

好吧，我也意识到你可以很容易地重新生成两个菜单：

import curses

def Main():
   x = 0
   while x!= ord('2'):           #draws 1st menu
       screen.clear();screen.border();
       screen.addstr(1,1, "Please choose:")
       screen.addstr(3,1, "1 - Another Menu")
       screen.addstr(4,1, "2 - Exit")
       x = screen.getch()         #grab input AFTER first giving options :)
       if x==ord('1'):            
           y = 0
           z = 0
           while y!= ord('2'):    #draws 2nd menu
               screen.clear();screen.border();
               screen.addstr(1,1, "Please choose from new menu:")
               screen.addstr(3,1, "1 - Do Something new")
               screen.addstr(4,1, "2 - Previous Menu")
               screen.addstr(6,1, "current loop : "+str(z))
               y = screen.getch();      #grabs new input
               while z!= -1:            #never breaks from loop unless 'break' is called
                   if y == ord('1'):
                       z += 1           
                       break   #regenerates 2nd menu
                   break   #regenerates 1st menu

           #Here we exit the internal loop.
           ##
##exit outside loop and close program
curses.endwin(); exit();

screen = curses.initscr()
Main()

Answer 1

在第二个循环结束后，添加 x = 0。

（你需要在每次循环开始时都把 x 重置为 0，而不是只在第一次的时候。如果不这样做，从第一个菜单退出时 x 会被设置为 "exit"，这样也会导致退出第二个菜单。）