Python 诅咒 : Returning to the Previous Menu

我在 ncurses 中有一个菜单系统。 选择其中一个选项会将您带到另一个菜单。但我该如何返回？

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，但我使用的设备没有内置该模块。

你们知道其他方法吗？非常感谢任何帮助。

============更新:==================

好的，我还意识到您可以轻松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() 

请您参考如下方法:

在第二个 while 循环结束后添加 x = 0。

(您需要在循环中每次重置x，而不仅仅是第一次。否则从第一个菜单退出将x设置“退出”，因此也将退出第二个菜单。)