Python字典实现简单的三级菜单(实例讲解)_Python

Python字典实现简单的三级菜单(实例讲解)

2020-11-29 00:33Python教程网 Python

下面小编就为大家带来一篇Python字典实现简单的三级菜单(实例讲解)。小编觉得挺不错的，现在就分享给大家，也给大家做个参考。一起跟随小编过来看看吧

如下所示：

									data = {

									 "北京":{

									  "昌平":{"沙河":["oldboy","test"],"天通苑":["链接地产","我爱我家"]},

									  "朝阳":{"望京":["奔驰","陌陌"],"国贸":["CICC","HP"],"东直门":["advent","飞信"]},

									  "海淀":{}

									 },

									 "山东":{

									  "德州":{},

									  "青岛":{},

									  "济南":{}

									 },

									 "广东":{

									  "东莞":{},

									  "常熟":{},

									  "佛山":{}

									 }

									}

									exit_flag = False

									while not exit_flag:

									 for i in data:

									  print(i)

									 choice_s = input("请输入以上任一省或直辖市名称>>>:")

									 if choice_s in data:

									  while not exit_flag:

									   for i in data[choice_s]:

									    print("\t",i)

									   choice_q = input("请输入以上任一市区名称>>>:")

									   if choice_q in data[choice_s]:

									    while not exit_flag:

									     for i in data[choice_s][choice_q]:

									      print("\t\t",i)

									     choice_j = input("请输入以上任一街道名称>>>:")

									     if choice_j in data[choice_s][choice_q]:

									      for i in data[choice_s][choice_q][choice_j]:

									       print("\t\t",i)

									      last = input("最后一层，按b返回>>>:")

									      if last == "b":

									       pass

									      elif last == "q":

									       exit_flag = True

									     if choice_j == "b":

									      break

									     elif choice_j == "q":

									      exit_flag = True

									   if choice_q == "b":

									    break

									   elif choice_q == "q":

									    exit_flag = True