python实现三级菜单
作者:互联网
多级菜单实现任务:
- 三级菜单
- 可依次选择进入各子菜单
- 所需新知识点:列表、字典
代码实现:
1 #Author:AZ 2 3 data = { 4 '北京':{ 5 "昌平":{ 6 "沙河":["old boy","test"], 7 "天通苑":["链家地产","我爱我家"] 8 }, 9 "朝阳":{ 10 "望京":["奔驰","陌陌"], 11 "国贸":["CICC","HP"], 12 "东直门":["Advent","飞信"] 13 }, 14 "海淀":[] 15 }, 16 '陕西':{ 17 "西安":{ 18 "雁塔区":["大雁塔"], 19 "曲江新区":["大唐芙蓉园"] 20 }, 21 "汉中":{ 22 "汉台区":["汉大","拜将坛"], 23 "洋县":["油菜花"] 24 }, 25 "宝鸡":[] 26 }, 27 '河南':{ 28 "郑州":{ 29 "二七":["银河世纪","人民公园"], 30 "金水区":["郑大"] 31 }, 32 "商丘":{ 33 "古城":["归德府","火神台"], 34 "高中":["二高","一高","四高"] 35 }, 36 "洛阳":{ 37 "洛龙区":["龙门石窟"], 38 "栾川县":["老君山"] 39 40 } 41 }, 42 43 } 44 exit_flag = False 45 46 while not exit_flag: 47 for i in data: 48 print(i) 49 choice = input("选择进入1>>>:") 50 if choice in data: 51 while not exit_flag: 52 for i2 in data[choice]: 53 print("\t",i2) 54 choice2 = input("选择输入2>>:") 55 if choice2 in data[choice]: 56 while not exit_flag: 57 for i3 in data[choice][choice2]: 58 print("\t\t",i3) 59 choice3 = input("选择进入3>>:") 60 if choice3 in data[choice][choice2]: 61 for i4 in data[choice][choice2][choice3]: 62 print("\t\t",i4) 63 choice4 = input("最后一层,按b返回>>:") 64 if choice4=="b": 65 pass 66 elif choice4=='q': 67 exit_flag=True 68 if choice3=='b': 69 break 70 elif choice3 =="q": 71 exit_flag=True 72 if choice2 =='b': 73 break 74 elif choice2=="q": 75 exit_flag = True3level_menus.py
标签:菜单,python,choice2,choice,flag,exit,choice3,data,三级 来源: https://www.cnblogs.com/dylan123/p/Dylan.html