输入1 可以持续添加数字，
输入2可以查看所有数字
输入数字0，退出系统
进行需求分析，编写简单的代码实现功能（本练习非常基础，有待完善）
import json
import os.path
import timemenu = """输入1 可以持续添加数字，
输入2可以查看所有数字
输入数字0，退出系统"""items = []
FILE_PATH = "./items.json"def save_data(): #存储with open(FILE_PATH, "w") as f:json.dump(items, f)def load_data():if os.path.exists(FILE_PATH): #加载with open(FILE_PATH, "r") as f:for data in json.load(f):items.append(data)def check_option():while True:print(menu)option = input("请输入选项")if option not in ["0", "1", "2"]:#简单判断用户输入print(f"输入不合法！重新输入")return optiondef add_num():while True:value = int(input("添加数字："))items.append(value)print(f"添加成功！")save_data()while True:select_yn = input("继续添加（y）,退至菜单（n）")#判断用户是否继续输入if select_yn not in ["y", "Y", "n", "N"]:print(f"输入不合法！")elif select_yn == "y" or select_yn == "Y":breakelif select_yn == "n" or select_yn == "N":breakif select_yn == "n" or select_yn == "N":breakdef look_all_num():if items: #进行简单的判断items是否为空for item in items:print(item)else:print(f"还没有添加数字")def quit_system():print(f"退出系统中。。。")time.sleep(2) #简单的系统休眠时间print(f"退出成功！")load_data()def main():while True:option = check_option()if option == "1":while True:add_num()elif option == "2":look_all_num()elif option == "0":quit_system()breakmain()
