最簡單的方式退出菜單

Question

我完成了我的所有作業，但我忘了在子菜單下放置一個取消選項。我不想更改所有文件，但需要2天時間才能完成。什麼將增加一個取消並返回到主菜單選項上sub_menu.py

這是sub_menu.py

#!/usr/bin/python 

## sub_menu.py 

#=========#=========#=========#=========#=========#=========#=========# 
def how_many(): 
    while True: 
     print 
     print " How many do you want to draw? -----Range is from 1 to 5-----" 
     shape_num = raw_input(' Enter your choice: ') 
     try: 
      shape_num = int(shape_num) 
      if (1 <= shape_num <= 5): 
       print " Number ", shape_num ," is OK." 
       break 
      else: 
       print 
       print " The number you entered must be from 1 to 5 only." 
     except: 
      print 
      print " Only numbers allowed - Please try again" 
    return shape_num 



#=========#=========#=========#=========#=========#=========#=========# 
def main(): 
    how_many() 
#=========#=========#=========#=========#=========#=========#=========# 
if __name__ == "__main__": 
    main() 

#=========#=========#=========#=========#=========#=========#=========# 

---

這是主要的一個

最簡單的方法

---

#!/usr/bin/python 

## my_menu.py 

import sub_menu as s_m 
import draw_shape as d_s 

#=========#=========#=========#=========#=========#=========#=========# 
def main(): 
    while True: 
     print 
     print " Draw a Shape" 
     print " ============" 
     print 
     print " 1 - Draw a triangle" 
     print " 2 - Draw a square" 
     print " 3 - Draw a rectangle" 
     print " 4 - Draw a pentagon" 
     print " 5 - Draw a hexagon" 
     print " 6 - Draw an octagon" 
     print " 7 - Draw a circle" 
     print 
     print " X - Exit" 
     print 

     choice = raw_input(' Enter your choice: ') 

     if (choice == 'x') or (choice == 'X'): 
      break 

     elif (choice >= "1" and choice <= "7"): 
      my_shape_num = s_m.how_many() 
      # draw in the middle of screen if == 1 
      if (my_shape_num == 1): 
       d_s.start_point(0, 0) 
      else: 
       d_s.start_point() 
      # 
      if choice == '1': 
       d_s.draw_triangle(my_shape_num) 
      elif choice == '2': 
       d_s.draw_square(my_shape_num) 
      elif choice == '3':    
       d_s.draw_rectangle(my_shape_num) 
      elif choice == '4':    
       d_s.draw_pentagon(my_shape_num) 
      elif choice == '5':    
       d_s.draw_hexagon(my_shape_num) 
      elif choice == '6':    
       d_s.draw_octagon(my_shape_num) 
      elif choice == '7': 
       d_s.draw_circle(my_shape_num) 

      d_s.t.end_fill() # shape fill color --draw_shape.py-- def start_point 

     else: 
      print 
      print ' Try again' 
      print 


#=========#=========#=========#=========#=========#=========#=========# 

if __name__ == "__main__": 
    main() 

#=========#=========#=========#=========#=========#=========#=========# 

---

這是一個drawshape

---

#!/usr/bin/python 

## draw_shape.py 

import turtle as t 

#=========#=========#=========#=========# 
def start_point(x=-240, y=0): 
    # Added 23 dec 11 baris 
    t.color("black", "red") 
    t.shapesize(outline=8) 
    t.begin_fill() 
    # 
    t.penup() 
    t.setpos(x, y) 
    t.pendown() 

def spaces_after(x=120): 
    t.penup() 
    t.forward(x) 
    t.pendown() 

#=========#=========#=========#=========# 
def draw_triangle(num=1): 
    for a in range(num): 
     for a in range(3): 
      t.forward(80) 
      t.left(120) 
     spaces_after() 
#=========#=========#=========#=========# 
def draw_square(num=1): 
    for a in range(num): 
     for a in range(4): 
      t.forward(80) 
      t.left(90) 
     spaces_after() 
#=========#=========#=========#=========#   
def draw_rectangle(num=1): 
    for a in range(num): 
     for a in range(2): 
      t.forward(80) 
      t.left(90) 
      t.forward(40) 
      t.left(90) 
     spaces_after() 
#=========#=========#=========#=========# 
def draw_pentagon(num=1): 
    for a in range(num): 
     for a in range(5): 
      t.forward(50) 
      t.left(72) 
     spaces_after() 
#=========#=========#=========#=========#   
def draw_hexagon(num=1): 
    for a in range(num): 
     for a in range(6): 
      t.forward(50) 
      t.left(60) 
     spaces_after() 
#=========#=========#=========#=========#   
def draw_octagon(num=1): 
    for a in range(num): 
     for a in range(8): 
      t.forward(40) 
      t.left(45) 
     spaces_after() 
#=========#=========#=========#=========# 
def draw_circle(num=1): 
    for a in range(num): 
     t . circle(50) 
     spaces_after() 
#=========#=========#=========#=========#=========#=========#=========# 
def main(): 
    start_point() 

    draw_triangle() 
    draw_square() 
    draw_rectangle() 
    draw_pentagon() 
    draw_hexagon() 
    draw_octagon() 
    draw_circle() 

    t.end_fill() 

#=========#=========#=========#=========#=========#=========#=========# 
if __name__ == "__main__": 
    main() 

#=========#=========#=========#=========#=========#=========#=========# 

Answer 1

你可以做到這一點在2行代碼。 在sub_menu.py你應該加 「0」 處理，像這樣

... 
try: 
      shape_num = int(shape_num) 
      if (shape_num == 0): 
       return None 
      if (1 <= shape_num <= 5): 
       print " Number ", shape_num ," is OK." 
       break 
      else: 
       print 
       print " The number you entered must be from 1 to 5 only." 
     except: 
.... 

和main.py你應該做

my_shape_num = s_m.how_many() 
      if (my_shape_num is None): 
       continue 

這應該重印主菜單。

如果需要，還應該添加說明（按0取消）。