我在ncurses中有一个菜单系统。 选择其中一个选项会转到另一个菜单。但我怎么回来?
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,但我正在使用的设备没有附带内置的模块。
你们知道其他任何方法吗?非常感谢任何帮助。
============更新:==================
好的,我也意识到你可以轻松地重新生成这两个菜单:
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()
答案 0 :(得分:1)
在第二个while循环结束后添加x = 0
。
(您需要在循环周围的每个时间重置x
,而不仅仅是第一个。否则退出第一个菜单x
将设置为“退出”,因此也将退出第二个菜单。)